Question

A Christmas light set consists of 40 identical bulbs in series across a 120 V line. If the total power used is 140 W, find the resistance of each bulb. 2.57 Omega 3.16 Omega 10.1 Omega 9.01 Omega Find the net charges of a system consisting of 6.51 * 10^6 electrons and 8.67 * 10^6 protons 1.46 10^-15 C 3.46 10^-13 C 2.46 10^13 C 5.46 10^-15 C A point charge q = +0.82 nC is fixed at the origin. At what distance a proton must be placed in order for the electric acting on the proton to be exactly opposite to its weight? 8480 m 7640 m 1011 m 5011 m (This problem relates to questions 19 and 20) Find the magnitude and direction (from the positive x axis) of the net electrostatic force exerted on the point charge q_2 in the figure below. Let q = +3.2 muC and d = 45 cm. Magnitude 3.41 N 7.61 N 1.01 N 4.03 N Direction 179 degree 195 degree 13.5 degree 265 degree What is the magnitude of the electric field produced by a charge of magnitude 5.01 muC at a distance of 1.65 m? 16500 N/C 1500 N/C 5500 N/C 13500 N/C A parallel-plate capacitor has plates separated by 0.58 mm. What is the potential difference between the plates if the electric field between the two plates has a magnitude of 3.2 * 10^4 V/m? 18.6 V 92.8 V 101 V 165 V A 0.49-muF capacitor is connected to a 12-V battery. What is the magnitude of the charge on each plate of the capacitor? 9.22 muC 5.88 muC 10.0 muC 1.65 muC

Answer 1

(8) (a) is correct.

In the series circuit, any current that flows through one bulb must go through the other bulbs as well, so each bulb draws the same current. Since all 40 bulbs have the same resistance, the voltage drop across each one is the same and equals (1/40)th of the applied voltage, or 120/40 = 3 volts.

Now, total power = 140 W implies,

$40\times I^{2} R=140$

$\Rightarrow 40\times \left (\frac{V}{40R} \right )^{2}\times R=140$

$\Rightarrow \frac{120^{2}}{40R}=140$

$\Rightarrow \boldsymbol{R=2.57\Omega}$

(9) Net charge is given by:

$q_{net}=n_{p}e-n_{e}e=(8.67\times10^{6}-6.51\times10^{6})\times1.602\times10^{-19}\ C=\boldsymbol{3.46\times10^{-13}\ C}$

Thus, (b) is correct.

(10) Force on proton = Weight of proton (W) = m_pg = (1.6726219 × 10^-27 x 9.81) N = 1.64 x 10^-26 N

Let x be the required distance. Then,

$F_{e}=W$

$\Rightarrow \frac{k_{e}qe}{x^{2}}=W$

$\Rightarrow \frac{9\times10^{9}\times0.82\times10^{-9}\times1.602\times10^{-19}}{x^{2}}=1.64\times10^{-26}$

Solving, we get x = 8480 m.

Thus (a) is correct.

(13) Electric field due to a point charge is given by:

$E_{5.01\mu C}=\frac{k_{e}q}{r^{2}}=\frac{9\times10^{9}\times5.01\times10^{-6}}{1.65^{2}}\approx \boldsymbol{16500\ N/C}$

Thus (a) is correct.

(14) The voltage difference between the two plates can be expressed in terms of the work done on a positive test charge q when it moves from the positive to the negative plate.

$V=\frac{work\ done}{charge}=\frac{Fd}{q}=Ed=3.2\times10^{4}\times0.58\times10^{-3}\ V=\boldsymbol{18.6\ V}$

Thus (a) is correct.

(15) Charge on capacitor plate is related to potential difference by following relation -

$Q=CV$

Therefore,

$Q_{plate}=0.49\times10^{-6}\times12=5.88\times10^{-6}C=\boldsymbol{5.88\mu C}$

Thus (b) is correct.