Electric potential due to a point charge of $\frac{1}{9} p C$ at a distance $1 k m$ from its centre is:

Question

Electric potential due to a point charge of dfrac-1-9- pC a distance 1k m from its centre is-0-1V10-6-V10-6V0V

Toppr Admin · Accepted Answer

Electric potential is given by-xA0-xA0-V-q4-x3C0-x3F5-or-kqr-xA0- -xA0-where k-14-x3C0-x3F5-o-9-xD7-109Hence- v-9-xD7-109-xD7-19-xD7-10-x2212-121000-x27F9-V-10-x2212-6VAnswer-B-

Electric potential due to a point charge of $\frac{1}{9} p C$ at a distance $1 k m$ from its centre is:
$0.1 V$
$10^{- 6} V$
$10^{6} V$
$0 V$

Electric potential is given by

$V = \frac{q}{4 π ϵ_{o} r} = \frac{k q}{r}$ where $k = \frac{1}{4 π ϵ_{o}} = 9 \times 10^{9}$

Hence, $v = \frac{9 \times 10^{9} \times \frac{1}{9} \times 10^{- 12}}{1000}$

$⟹ V = 10^{- 6} V$

Answer-(B)

Electric potential due to a point charge of 19pC at a distance 1km from its centre is:0.1V10−6V106V0V

Electric potential is given by V=q4πϵor=kqr where k=14πϵo=9×109Hence, v=9×109×19×10−121000⟹V=10−6VAnswer-(B)

Electric potential due to a point charge of $\frac{1}{9} p C$ at a distance $1 k m$ from its centre is:
$0.1 V$
$10^{- 6} V$
$10^{6} V$
$0 V$

Electric potential is given by

$V = \frac{q}{4 π ϵ_{o} r} = \frac{k q}{r}$ where $k = \frac{1}{4 π ϵ_{o}} = 9 \times 10^{9}$

Hence, $v = \frac{9 \times 10^{9} \times \frac{1}{9} \times 10^{- 12}}{1000}$

$⟹ V = 10^{- 6} V$

Answer-(B)