Physics Problems & Solutions: Method of Image

Showing posts with label Method of Image. Show all posts

Electromagnetism - Method of Image

Two positive charged of q and 2q coulombs are located on the x-axis at x = 0.5a and 1.5a respectively, as shown above. There is an infinite grounded conducting plane at x = 0. What is the magnitude of the net force on the charge q?

A. $\frac{1}{4\pi \epsilon_0} \frac{q^2}{a^2}$

B. $\frac{1}{4\pi \epsilon_0} \frac{3q^2}{2a^2}$

C. $\frac{1}{4\pi \epsilon_0} \frac{2q^2}{a^2}$

D. $\frac{1}{4\pi \epsilon_0} \frac{3q^2}{a^2}$

E. $\frac{1}{4\pi \epsilon_0} \frac{7q^2}{2a^2}$

(GR9277 #10)

Solution:

Using method of image:

On q:

$\begin{aligned} \vec{F}_{\text{total}}&=kq\left [ \frac{2q}{a^2}\left ( -\hat{x} \right )+ \frac{-q}{a^2}\hat{x} +\frac{-2q}{(2a)^2}\hat{x} \right ] \\&=\frac{kq^2}{a^2}\left ( -2-1-\frac{1}{2} \right )\hat{x} \\&=-\frac{7}{2}\frac{kq^2}{a^2}\hat{x} \\\left |\vec{F}_{\text{total}} \right |&=k\frac{7q^2}{2a^2}\end{aligned}$

with $k=\frac{1}{4\pi \epsilon_0}$

Answer: E

Electromagnetism - Method of Image

A positive charge Q is located at a distance L above an infinite grounded conducting plane, as shown in the figure above. What is the total charge induced on the plane?

A. 2Q
B. Q
C. 0
D. –Q
E. –2Q

(GR0177 #08)

Solution:

Method of image:

+Q attracts negative charges in the plate and repels positive ones.
As the positive charges want to “get away”, they succeed in doing so through the ground, leaving the negative charges behind.
The plate is left with a net negative charge.

Answer: D

Subscribe to: Posts ( Atom )