Two point charges -q and -q are located points -0-0-a- and -0-0-a- respectively- The potential a point -0-0-z- where z-a isdfrac-2qa-4piepsilon-0 -z-2-a-2-dfrac-q-4piepsilon-0a-dfrac-qa-4piepsilon-0 z-2-dfrac-2qa-4piepsilon-0 -z-2-a-2-

Question

Two point charges -q and -q are located points -0-0-a- and -0-0-a- respectively- The potential a point -0-0-z- where z-a  isdfrac-2qa-4piepsilon-0 -z-2-a-2-dfrac-q-4piepsilon-0a-dfrac-qa-4piepsilon-0 z-2-dfrac-2qa-4piepsilon-0 -z-2-a-2-

Toppr Admin · Accepted Answer

The correct option is D 2qa4-x3C0-x3F5-0-z2-x2212-a2-V-V1-V2-x27F9-V-q4-x3C0-x3F5-o-z-a-x2212-q4-x3C0-x3F5-o-z-x2212-a-x27F9-V-2qa4-x3C0-x3F5-o-z2-x2212-a2-Answer-D-

Two point charges $+ q$ and $- q$ are located at points $(0, 0, - a)$ and $(0, 0, a)$ respectively. The potential at a point $(0, 0, z)$ where $z > a$ is
$\frac{2 q a}{4 π ϵ_{0} (z^{2} + a^{2})}$
$\frac{q}{4 π ϵ_{0} a}$
$\frac{q a}{4 π ϵ_{0} z^{2}}$
$\frac{2 q a}{4 π ϵ_{0} (z^{2} - a^{2})}$

The correct option is D $\frac{2 q a}{4 π ϵ_{0} (z^{2} - a^{2})}$

$V = V_{1} + V_{2}$

$⟹ V = \frac{q}{4 π ϵ_{o} (z + a)} - \frac{q}{4 π ϵ_{o} (z - a)}$

$⟹ V = \frac{2 q a}{4 π ϵ_{o} (z^{2} - a^{2})}$

Answer-(D)

Two point charges +q and −q are located at points (0,0,−a) and (0,0,a) respectively. The potential at a point (0,0,z) where z>a is2qa4πϵ0(z2+a2)q4πϵ0aqa4πϵ0z22qa4πϵ0(z2−a2)

The correct option is D 2qa4πϵ0(z2−a2)V=V1+V2⟹V=q4πϵo(z+a)−q4πϵo(z−a)⟹V=2qa4πϵo(z2−a2)Answer-(D)

Two point charges $+ q$ and $- q$ are located at points $(0, 0, - a)$ and $(0, 0, a)$ respectively. The potential at a point $(0, 0, z)$ where $z > a$ is
$\frac{2 q a}{4 π ϵ_{0} (z^{2} + a^{2})}$
$\frac{q}{4 π ϵ_{0} a}$
$\frac{q a}{4 π ϵ_{0} z^{2}}$
$\frac{2 q a}{4 π ϵ_{0} (z^{2} - a^{2})}$

The correct option is D $\frac{2 q a}{4 π ϵ_{0} (z^{2} - a^{2})}$

$V = V_{1} + V_{2}$

$⟹ V = \frac{q}{4 π ϵ_{o} (z + a)} - \frac{q}{4 π ϵ_{o} (z - a)}$

$⟹ V = \frac{2 q a}{4 π ϵ_{o} (z^{2} - a^{2})}$

Answer-(D)