Question

Four-point charges Q, q, Q and q are placed at the corners of a square of side ‘a’ as shown in the figure.


Find the

(a) resultant electric force on a charge Q, and

(b) potential energy of this system.

Answer 1

Answer:

a) $F_{q} = \frac{Q}{4\pi\varepsilon _{0} } *\frac{1}{a^{2} } [ 2q +\frac{Q}{2} ]$

b) $PE = -\frac{1}{4\pi \varepsilon_{0} a} [ 4Qq +\frac{q^{2} }{\sqrt{2} } +\frac{Q^{2} }{\sqrt{2} } ]$

Explanation:

I have attached the image for reference of my solution.

a) resultant electric force on a charge Q.

The length of diagonal = a√2

The resultant electric force on charge Q can be given by $F_{q}$

$F_{q} = \frac{kqQ}{a^{2} } + \frac{kqQ}{a^{2} } +\frac{kQ^{2} }{(a\sqrt{2})^{2} }$

$F_{q} = \frac{Q}{4\pi\varepsilon _{0} } *\frac{1}{a^{2} } [ 2q +\frac{Q}{2} ]$

Hence this the required resultant force on a charge Q,

b) Potential energy of this system.

$PE_{system} = -w$

$w_{ \infty} \to A =0$

$w_{ \infty} \to B =\frac{kQq}{a}$

$w_{ \infty} \to D =\frac{kQq}{a}+\frac{kq^{2} }{a\sqrt{2} }$

$PE = - ( w_{ \infty} \to A + w_{ \infty} \to B+w_{ \infty} \to C+w_{ \infty} \to D )$

$PE = - ( 0 + \frac{kQq}{a} +\frac{kQq}{a} +\frac{kq^{2} }{a\sqrt{2} } +\frac{kQ^{2} }{a\sqrt{2} } +\frac{2kqQ}{a}$

$PE = -\frac{1}{4\pi \varepsilon_{0} a} [ 4Qq +\frac{q^{2} }{\sqrt{2} } +\frac{Q^{2} }{\sqrt{2} } ]$

Hence this the required potential energy of this system.