Answer to Question #186792 in Electric Circuits for Muzala Seleji

Question

1. A 35 𝜇𝐶 point charge is placed 32 𝑐𝑚 from an identical 32 𝜇𝐶 charge. How much work would be required to move a 0.50 𝜇𝐶 charge from a point midway between them to a point 12 𝑐𝑚 closer to either of the charges?

2. Three charges are at the vertices of an isosceles triangle. With 𝑞=7.00 𝑛𝐶, the two charges at the base of the triangle are separated by 2.00 𝑐𝑚 and each have a charge of −𝑞. The charge q at the top of the triangle is 4.00 𝑐𝑚 from each of the charges at the base. Calculate the electric potential halfway between the charges on the base of the triangle.

3. (a) Show that the capacitance of a parallel-plate capacitor is given by the equation

C=EA/D

(b) A 1-megabit computer memory chip contains many 60.0×10−15 𝐹 capacitors. Each capacitor has a plate area of 21.0×10−12 𝑚2. Determine the plate separation of such a capacitor. (Assume a parallel-plate configuration.) The diameter of atom is on the order of 10−10 𝑚 = 1 ˚ 𝐴. Express the plate separation in angstroms.

Answer 1

(1) $q_1=q_2=q=35\mu C$

$q_3=0.50\mu C$

Work Done $=qV=q_3(V_b-V_a)$

$(0.50\mu C)(k\times35\mu C)\Bigg[\left(\dfrac{1}{0.04\space m}+\dfrac{1}{0.28\space m}\right)-\left(\dfrac{1}{0.16\space m}+\dfrac{1}{0.16\space m}\right)\Bigg]$

$=2.53\space J$

(2) $V_1=\dfrac{kq}{r_1}=\dfrac{9\times10^9\times7\times10^{-9}}{\sqrt{15}\times10^{-2}}=1626.65\space V$

$V_2=\dfrac{k(-q)}{r_2}=\dfrac{9\times10^9\times(-7)\times10^{-9}}{{1}\times10^{-2}}=-6300\space V$

$V_3=\dfrac{k(-q)}{r_3}=\dfrac{9\times10^9\times(-7)\times10^{-9}}{{1}\times10^{-2}}=-6300\space V$

$\sum V=V_1+V_2+V_3=-10973.35\space V$

(3)

The flux through the curved part outside the plates is also zero as the direction of the field E is parallel to this surface. The flux through $\Delta A$ is

$\Phi=\vec{E}.\Delta \vec A=E\Delta A$

The only charge inside the Gaussian surface is

$\Delta Q=\sigma\Delta A=\dfrac{Q}{A}\Delta A$

From Gauss’s law,

$\oint\vec E.d\vec S=\dfrac{Q_{in}}{\epsilon_o}$

$E\Delta A=\dfrac{Q}{\epsilon_oA}\Delta A$

$E=\dfrac{Q}{\epsilon_o A}$

The potential difference between the plates is

$V=V_+-V_-=-\int_{A}^{B} \vec E \,.d\vec r$

$V=\int_A^BEdr$

$V=Ed=\dfrac{Qd}{\epsilon_o A}$

The capacitance of the parallel-plate capacitor is

$C=\dfrac{Q}{V}=\dfrac{Q\epsilon_o A}{Qd}$

$C=\dfrac{\epsilon_o A}{d}$

(b) $C=60\times10^{-15}\space F$

Area, $A=21\times10^{-12}\space m^2$

$d=\dfrac{\epsilon_o A}{C}=\dfrac{8.85\times10^{-12}\times21\times10^{-12}}{60\times10^{-15}}$

$d=3.0975\times10^{-9}\space m=30.975\space A^o$