Question 812: Electrostats & Capacitance

Question

Two point charges $(A$ and $B)+4 q$ and $-4 q$ are placed along a line separated by a distance I '. Force acting between them is F. If $25 \%$ of charge from point $A$ is transferred to that at point B , the force between the charges now becomes

Accepted Answer

When the charges $+4q$ and $-4q$ are placed along a line separated by a distance $ r $, the initial force between them is given by:
$ F = \frac{-16q^2}{4\pi\varepsilon_0 r^2} \quad 	ext{..... (i)} $
If $25\%$ of the charge from point $A$ is transferred to point $B$, the charges change as follows:
Charge at point $A$ becomes:
$   q_1 = +4q - 0.25(+4q) = +3q   $
Charge at point $B$ becomes:
$   q_2 = -4q + 0.25(+4q) = -3q   $
Thus, the new force between points $A$ and $B$ is:
$ F' = \frac{(3q) 	imes (-3q)}{4\pi\varepsilon_0 r^2} = \frac{-9q^2}{4\pi\varepsilon_0 r^2} $
By multiplying and dividing by 16 to compare with the initial force $F$, we get:
$ F' = \frac{9}{16} 	imes \left(\frac{-16q^2}{4\pi\varepsilon_0 r^2}ight) = \frac{9}{16} F \quad [	ext{From (i)}] $

Answer

$\frac{3}{4} \mathrm{~F}$

Answer

$\frac{4}{3} \mathrm{~F}$

Answer

$\frac{9}{16} \mathrm{~F}$

Answer

$\frac{16}{9} \mathrm{~F}$

Electrostats & Capacitance

Two point charges (A and B)+4 q and -4 q are placed along a line separated by a distance I '. Force acting between them is F. If 25 \% of charge from point A is transferred to that at point B , the force between the charges now becomes

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