Two charges are at a distance d apart. If a copper plate of thickness (d/2) is kept between them, the effective force will be

Question

Two charges are at a distance d apart. If a copper plate of thickness (d/2) is kept between them, the effective force will be (A)  (F/2)  (B) zero (C)  2F  (D)  √2F

Tardigrade Admin · Accepted Answer

From Coulomb's law, electric force between two charges is directly proportional to product of charges and inversely proportional to square of distance between them. That is

F = k \frac{q _{1} q _{2}}{d ^{2}}

where,

k = \frac{1}{4 π ε _{0}} =

proportionality constant. If a medium is placed between the charges, then

F^{'} = \frac{1}{4 π ε _{0} K} \frac{q _{1} q _{2}}{d ^{2}}

Since, medium placed between the charges is a metallic plate, so for it

K = \infty

Hence,

F^{'} = 0

(zero)

Q. Two charges are at a distance $d$ apart. If a copper plate of thickness $\frac{d}{2}$ is kept between them, the effective force will be

$\frac{F}{2}$

zero

$2 F$

$2 F$

Q. Two charges are at a distance d apart. If a copper plate of thickness 2d​ is kept between them, the effective force will be

2F​

zero

2F

2F​

Q. Two charges are at a distance $d$ apart. If a copper plate of thickness $\frac{d}{2}$ is kept between them, the effective force will be

$\frac{F}{2}$

$2 F$

$2 F$