Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by ρ (r) = kr, where r is the distance from the centre. Two charges A and B, of -Q each, are placed on diametrically opposite points, at equal distance, a, from the centre. If A and B do not experience any force, then :

Question

Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by ρ (r) = kr, where r is the distance from the centre. Two charges A and B, of -Q each, are placed on diametrically opposite points, at equal distance, a, from the centre. If A and B do not experience any force, then : (A) a = (3R/21/4) (B) a = R / √3 (C) a = 8-1/4 R (D) a = 2-1/4 R

Tardigrade Admin · Accepted Answer

E 4π a2 = (∫0a kr 4 π r2 dr/ϵ) E = (k 4 π a4/4 × 4 π ϵ0) 2Q = ∫0R kr 4 π r2 dr k = (2Q/π R4) QE = (1/4 π ϵ0) (QQ/(2a)2) R = a81/4

$a = \frac{3 R}{2 ^{1/4}}$

$a = R / 3$

$a = 8^{- 1/4 R}$

$a = 2^{- 1/4} R$

$E 4 π a^{2} = \frac{\int _{0}^{a} k r 4 π r ^{2} d r}{ϵ}$
$E = \frac{k 4 π a ^{4}}{4 \times 4 π ϵ _{0}}$
$2 Q = \int_{0}^{R} k r 4 π r^{2} d r$
$k = \frac{2 Q}{π R ^{4}}$
$QE = \frac{1}{4 π ϵ _{0}} \frac{QQ}{( 2 a ) ^{2}}$
$R = a 8^{1/4}$

a=21/43R​

a=R/3​

a=8−1/4R

a=2−1/4R

E4πa2=ϵ∫0a​kr4πr2dr​ E=4×4πϵ0​k4πa4​ 2Q=∫0R​kr4πr2dr k=πR42Q​ QE=4πϵ0​1​(2a)2QQ​ R=a81/4

$a = \frac{3 R}{2 ^{1/4}}$

$a = R / 3$

$a = 8^{- 1/4 R}$

$a = 2^{- 1/4} R$

$E 4 π a^{2} = \frac{\int _{0}^{a} k r 4 π r ^{2} d r}{ϵ}$
$E = \frac{k 4 π a ^{4}}{4 \times 4 π ϵ _{0}}$
$2 Q = \int_{0}^{R} k r 4 π r^{2} d r$
$k = \frac{2 Q}{π R ^{4}}$
$QE = \frac{1}{4 π ϵ _{0}} \frac{QQ}{( 2 a ) ^{2}}$
$R = a 8^{1/4}$