Two equal point charges are fixed at x =- a and x = + a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to

Question

Two equal point charges are fixed at x =- a and x = + a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to (A) x (B) x2  (C) x3  (D) 1/x

Tardigrade Admin · Accepted Answer

Ui= (2KQq/a)+ (K.q.q./2a) and Uf=KQq [ (1/a+x)+ (1/a-x) ]+ (K.q.q./2a) Here, K= (1/4 π ε0) Δ U=Uf-Ui or |Δ U|= (2KQqx2/a3) For x < < a ∴ Δ Ux2

Q. Two equal point charges are fixed at x =- a and x = + a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to

x

$x^{2}$

$x^{3}$

$1/ x$

$U_{i} = \frac{2 K Qq}{a} + \frac{K . q . q .}{2 a}$ and $U_{f} = K Q_{q} [\frac{1}{a + x} + \frac{1}{a - x}] + \frac{K . q . q .}{2 a}$ Here, $K = \frac{1}{4 π ε _{0}}$ $Δ U = U_{f} - U_{i}$ or $∣Δ U ∣ = \frac{2 K Qq x ^{2}}{a ^{3}}$ $F or x << a$ $∴ Δ U x^{2}$

Q. Two equal point charges are fixed at x =- a and x = + a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to

x

x2

x3

1/x

Ui​=a2KQq​+2aK.q.q.​ and Uf​=KQq​[a+x1​+a−x1​]+2aK.q.q.​ Here, K=4πε0​1​ ΔU=Uf​−Ui​ or ∣ΔU∣=a32KQqx2​ Forx<<a ∴ΔUx2

$x^{2}$

$x^{3}$

$1/ x$

$U_{i} = \frac{2 K Qq}{a} + \frac{K . q . q .}{2 a}$ and $U_{f} = K Q_{q} [\frac{1}{a + x} + \frac{1}{a - x}] + \frac{K . q . q .}{2 a}$ Here, $K = \frac{1}{4 π ε _{0}}$ $Δ U = U_{f} - U_{i}$ or $∣Δ U ∣ = \frac{2 K Qq x ^{2}}{a ^{3}}$ $F or x << a$ $∴ Δ U x^{2}$