Practice Exam 1

Electric Charge & Electric Fields

Questions10

Time90 min

Points100

CalculatorAllowed

Use \(k = 8.99\times10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2\), \(\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/(\text{N}\cdot\text{m}^2)\), \(e = 1.60\times10^{-19}\) C, \(m_e = 9.11\times10^{-31}\) kg, \(g = 9.8\ \text{m/s}^2\).

1Conductors & induction8 pts

A positively charged glass rod is brought near (not touching) a neutral metal sphere on an insulating stand. (a) Is the sphere attracted, repelled, or unaffected? Explain. (b) Describe how to leave the sphere with a net negative charge without touching it with the rod. (c) Which particles actually move in (b)?

Solution

(a) Attracted. The rod pulls free electrons in the metal toward the near side, leaving the far side positive. The sphere stays neutral overall, but the negative side is closer to the rod. Since Coulomb force grows as \(1/r^2\), the attraction of the near side beats the repulsion of the far side.

(b) Charging by induction: with the rod held near, connect the sphere to ground (touch it with a grounded wire). Electrons flow up from ground, attracted by the rod. Break the ground connection while the rod is still there, then take the rod away. The extra electrons are trapped, and they spread over the sphere: net charge negative, opposite to the rod.

(c) Only electrons move, from the ground into the sphere. The positive nuclei stay where they are. The rod loses no charge at all.

2Quantization8 pts

A small metal sphere is given a charge of \(+3.2\) nC. (a) How many electrons were removed? (b) By how much did its mass change? (c) Could an object ever carry a charge of \(2.4\times10^{-19}\) C? Explain.

Solution

\[ n = \frac{q}{e} = \frac{3.2\times10^{-9}}{1.60\times10^{-19}} = \mathbf{2.0\times10^{10}}\ \text{electrons removed} \] \[ \Delta m = n\,m_e = (2.0\times10^{10})(9.11\times10^{-31}) = \mathbf{1.8\times10^{-20}\ kg}\ \text{lighter} \]

Positive means electrons were removed, so the mass goes down — though by far too little to measure.

(c) No. \(2.4\times10^{-19}/1.60\times10^{-19} = 1.5\). Free charge comes only in whole multiples of \(e\), so 1.5\(e\) is impossible.

3Coulomb's law in vector form10 pts

\(q_1 = +4.0\ \mu\text{C}\) is at the origin and \(q_2 = -2.0\ \mu\text{C}\) is at \((3.0, 4.0)\) cm. Find the force on \(q_2\) in component form, and its magnitude.

Solution

Distance \(r = \sqrt{3^2+4^2} = 5.0\) cm \(= 0.050\) m. Unit vector from \(q_1\) to \(q_2\): \(\hat{\mathbf r} = (0.6, 0.8)\).

\[ \mathbf F = k\frac{q_1q_2}{r^2}\hat{\mathbf r} = \frac{(8.99\times10^{9})(4.0\times10^{-6})(-2.0\times10^{-6})}{(0.050)^2}(0.6,\ 0.8) = (-28.8\ \text{N})(0.6,\ 0.8) \] \[ \boxed{\mathbf F = (-17.3,\ -23.0)\ \text{N}},\qquad |\mathbf F| = 28.8\ \text{N} \]

The product \(q_1q_2\) is negative, so the force points along \(-\hat{\mathbf r}\): back toward the origin. Unlike charges attract ✓. Note how large this is — about the weight of a 3 kg mass — from just a few microcoulombs 5 cm apart.

4Where is the field zero?10 pts

A charge \(+4q\) is at \(x = 0\) and a charge \(+q\) is at \(x = 0.30\) m. (a) Find the point on the \(x\)-axis where \(\mathbf E = 0\). (b) Repeat if the second charge is \(-q\) instead.

Solution

(a) Both positive: their fields point in opposite directions only between them. Let the point be at \(x\), \(0<x<0.30\):

\[ \frac{k(4q)}{x^2} = \frac{kq}{(0.30-x)^2}\ \Rightarrow\ \frac{2}{x} = \frac{1}{0.30-x}\ \Rightarrow\ 0.60 - 2x = x\ \Rightarrow\ \boxed{x = 0.20\ \text{m}} \]

(Taking the square root keeps only the positive roots, which is right here since both distances are positive.) The zero is closer to the smaller charge, as it must be.

(b) Opposite signs: between them both fields point the same way (toward \(-q\)), so no cancellation there. The zero must be outside, on the side of the smaller charge, \(x > 0.30\):

\[ \frac{k(4q)}{x^2} = \frac{kq}{(x-0.30)^2}\ \Rightarrow\ 2(x - 0.30) = x\ \Rightarrow\ \boxed{x = 0.60\ \text{m}} \]

Check: at 0.60 m, \(4q/0.36 = 11.1q\) and \(q/0.09 = 11.1q\) ✓, pointing in opposite directions.

5Superposition with symmetry10 pts

Three charges of \(+1.0\) nC sit at the corners of an equilateral triangle of side 10 cm. (a) What is \(\mathbf E\) at the centre? (b) One of the charges is replaced by \(-1.0\) nC. Find \(\mathbf E\) at the centre now. (Hint: avoid computing components.)

Solution

(a) Three equal fields at \(120^\circ\) to each other: by symmetry, \(\mathbf E = \mathbf 0\).

(b) Replacing \(+q\) by \(-q\) at corner A is the same as adding \(-2q\) at A to the original (zero-field) arrangement. So the new field is just the field of \(-2q\) at A. Distance from a corner to the centre: \(r = s/\sqrt3 = 0.10/\sqrt3 = 0.0577\) m.

\[ E = \frac{k(2q)}{r^2} = \frac{2(8.99\times10^{9})(1.0\times10^{-9})}{(0.0577)^2} = \frac{17.98}{3.33\times10^{-3}} = \boxed{5.39\times10^{3}\ \text{N/C}} \]

Direction: toward the negative charge (field of a negative charge points toward it). Superposition lets you add and subtract whole arrangements, not just individual charges.

6Continuous distribution: derivation12 pts

A thin rod bent into a semicircle of radius \(R\) carries charge \(+Q\) uniformly. (a) Set up and evaluate the integral for \(\mathbf E\) at the centre of the circle. (b) Evaluate for \(Q = 5.0\) nC, \(R = 10\) cm.

Solution

Put the centre at the origin and the arc in the upper half-plane, \(0\le\theta\le\pi\). Linear density \(\lambda = Q/(\pi R)\). A piece at angle \(\theta\) has length \(R\,d\theta\), so \(dq = \lambda R\,d\theta\), and it is a distance \(R\) from the centre.

Symmetry: pieces at \(\theta\) and \(\pi-\theta\) have opposite \(x\)-components, which cancel. Each \(d\mathbf E\) points away from its piece, so the surviving component points in \(-y\), with fraction \(\sin\theta\):

\[ E_y = -\int_0^{\pi}\frac{k\,\lambda R\,d\theta}{R^2}\sin\theta = -\frac{k\lambda}{R}\Big[-\cos\theta\Big]_0^{\pi} = -\frac{2k\lambda}{R} = -\frac{2kQ}{\pi R^2} \]

(b)

\[ E = \frac{2(8.99\times10^{9})(5.0\times10^{-9})}{\pi(0.10)^2} = \frac{89.9}{0.0314} = \boxed{2.86\times10^{3}\ \text{N/C}} \]

pointing away from the arc (toward the open side). Sanity check: a point charge \(Q\) at distance \(R\) would give \(kQ/R^2 = 4.5\times10^{3}\) N/C; the arc gives less (factor \(2/\pi\)) because its pieces pull in partly cancelling directions.

7Ring of charge10 pts

A ring of radius \(a\) carries total charge \(Q\). Its axial field is \(E(z) = kQz/(z^2+a^2)^{3/2}\). (a) Explain why \(E(0) = 0\). (b) Show that the maximum is at \(z = a/\sqrt2\) and find \(E_{\max}\) in terms of \(kQ/a^2\). (c) Show that for \(z\gg a\) the ring acts like a point charge.

Solution

(a) At the centre every piece \(dq\) is matched by an identical piece directly opposite, whose field is equal and opposite. Everything cancels.

(b)

\[ \frac{dE}{dz} = kQ\,\frac{(z^2+a^2)^{3/2} - z\cdot\tfrac32(z^2+a^2)^{1/2}(2z)}{(z^2+a^2)^{3}} = kQ\,\frac{a^2 - 2z^2}{(z^2+a^2)^{5/2}} = 0\ \Rightarrow\ z = \frac{a}{\sqrt2} \] \[ E_{\max} = \frac{kQ\,(a/\sqrt2)}{(3a^2/2)^{3/2}} = \frac{2}{3\sqrt3}\,\frac{kQ}{a^2} \approx \boxed{0.385\,\frac{kQ}{a^2}} \]

(c) For \(z\gg a\), \(z^2 + a^2\approx z^2\), so \(E\approx kQz/z^3 = kQ/z^2\) — Coulomb's law for a point charge \(Q\) ✓.

8Dipole: torque, energy, far field10 pts

A dipole consists of \(\pm1.5\) nC separated by 4.0 mm. It is in a uniform field of \(3.0\times10^{4}\) N/C, with \(\mathbf p\) at \(60^\circ\) to \(\mathbf E\). Find (a) \(p\); (b) the torque; (c) the potential energy; (d) the work an external agent must do to rotate the dipole from \(\theta = 0^\circ\) to \(180^\circ\); (e) the dipole's own field on its axis 20 cm away.

Solution

\[ \text{(a)}\ p = qd = (1.5\times10^{-9})(4.0\times10^{-3}) = 6.0\times10^{-12}\ \text{C}\cdot\text{m} \] \[ \text{(b)}\ \tau = pE\sin60^\circ = (6.0\times10^{-12})(3.0\times10^{4})(0.866) = 1.56\times10^{-7}\ \text{N}\cdot\text{m} \] \[ \text{(c)}\ U = -pE\cos60^\circ = -(1.8\times10^{-7})(0.500) = -9.0\times10^{-8}\ \text{J} \] \[ \text{(d)}\ W = U(180^\circ) - U(0^\circ) = 2pE = 3.6\times10^{-7}\ \text{J} \] \[ \text{(e)}\ E = \frac{2kp}{r^3} = \frac{2(8.99\times10^{9})(6.0\times10^{-12})}{(0.20)^3} = 13.5\ \text{N/C} \]

The far-field formula is fine since \(r = 200\) mm \(\gg d = 4\) mm. The torque in (b) acts to rotate \(\mathbf p\) toward \(\mathbf E\), reducing \(\theta\).

9EE application: inkjet deflection12 pts

In a continuous inkjet printer, an ink drop of mass \(1.3\times10^{-10}\) kg and charge \(-1.5\times10^{-13}\) C moves horizontally at 20 m/s into a region 2.0 cm long between deflection plates, where \(\mathbf E = 8.0\times10^{5}\) N/C points up. (a) Find the vertical deflection as it leaves the plates, ignoring gravity. (b) Is ignoring gravity justified? (c) The printer controls where the drop lands by choosing its charge. What charge would give a deflection of 1.0 mm?

Solution

\[ a = \frac{|q|E}{m} = \frac{(1.5\times10^{-13})(8.0\times10^{5})}{1.3\times10^{-10}} = 923\ \text{m/s}^2\ \text{(down: negative charge, field up)} \] \[ t = \frac{L}{v_0} = \frac{0.020}{20} = 1.0\times10^{-3}\ \text{s},\qquad y = \tfrac12at^2 = \tfrac12(923)(1.0\times10^{-6}) = \boxed{4.6\times10^{-4}\ \text{m} = 0.46\ \text{mm}} \]

(b) \(g = 9.8\ \text{m/s}^2\) is about 1% of 923 m/s². Including it gives 0.47 mm. Not negligible to the same degree as for an electron, but small. (For a drop, always compare; for an electron, never bother.)

(c) \(y\propto |q|\), so a 1.0 mm deflection needs \(1.0/0.46 = 2.17\) times the charge: \(|q|\approx 3.3\times10^{-13}\) C (negative). This is exactly how the printer works — a charging electrode gives each drop a controlled charge, and fixed plates do the steering.

10Stopping an electron10 pts

An electron moving in the \(+x\) direction at \(3.0\times10^{6}\) m/s enters a region with a uniform field \(E = 500\) N/C, also in the \(+x\) direction. (a) How far does it travel before stopping? (b) How long does that take? (c) What happens next?

Solution

The force on the electron is \(-e\mathbf E\): in \(-x\), opposite its velocity, so it slows down.

\[ a = \frac{eE}{m_e} = \frac{(1.60\times10^{-19})(500)}{9.11\times10^{-31}} = 8.78\times10^{13}\ \text{m/s}^2 \] \[ \text{(a)}\ d = \frac{v_0^2}{2a} = \frac{(3.0\times10^{6})^2}{2(8.78\times10^{13})} = \boxed{5.1\times10^{-2}\ \text{m} = 5.1\ \text{cm}} \] \[ \text{(b)}\ t = \frac{v_0}{a} = \frac{3.0\times10^{6}}{8.78\times10^{13}} = \boxed{3.4\times10^{-8}\ \text{s} = 34\ \text{ns}} \]

(c) The force does not stop when the velocity does. The electron accelerates back in \(-x\) and leaves the region where it entered, at \(3.0\times10^{6}\) m/s (the motion is symmetric, like a ball thrown upward). A common error is to assume the electron is pushed along \(\mathbf E\) — it is pushed against it.

Scoring

Score	Where you are
90–100	Solid. On to Gauss's law.
75–89	Ideas fine; slips in signs or unit conversions (cm to m, nC to C). Re-check the negative-charge questions (Q9, Q10) and the direction rules in Section 4.
60–74	Vector addition or integral set-up shaky (Q3, Q5, Q6). Re-read Section 3 and Section 6 with the lab open; redo the ring by hand.
< 60	Start from Section 2: Coulomb's law with directions must be automatic before fields will make sense. Then dipoles and motion.

Where the points usually go

Adding field magnitudes instead of components (Q5, Q6); putting the zero-field point on the wrong side for opposite charges (Q4b); forgetting that an electron is pushed against \(\mathbf E\) (Q9, Q10); and leaving \(r\) in centimetres (Q3). In Q6, forgetting that only the \(y\)-components survive gives \(kQ/R^2\) — wrong by \(\pi/2\).