What is the electric potential a distance of 9 cm from 3 nC-270 V3 V300 V30 V

Question

Toppr Admin · Accepted Answer

Given - -xA0- -xA0-r-9 cm -0-09 m -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0-Charge -xA0- -xA0-Q-3 nC -3-xD7-10-x2212-9 CPotential at a distance 9 cm- -xA0- -xA0- -xA0-V-kQr -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0-where k-9-xD7-109-x2234- -xA0- V-9-xD7-109-xD7-3-xD7-10-x2212-90-09-300 V

What is the electric potential at a distance of 9 cm from 3 nC?
$270 V$
$3 V$
$300 V$
$30 V$

Given : $r = 9$ cm $= 0.09$ m
Charge $Q = 3$ nC $= 3 \times 10^{- 9}$ C
Potential at a distance 9 cm, $V = \frac{k Q}{r}$ where $k = 9 \times 10^{9}$
$∴$ $V = \frac{9 \times 10^{9} \times 3 \times 10^{- 9}}{0.09} = 300$ V

What is the electric potential at a distance of 9 cm from 3 nC?270V3V300V30V

Given : r=9 cm =0.09 m Charge Q=3 nC =3×10−9 CPotential at a distance 9 cm, V=kQr where k=9×109∴ V=9×109×3×10−90.09=300 V

What is the electric potential at a distance of 9 cm from 3 nC?
$270 V$
$3 V$
$300 V$
$30 V$

Given : $r = 9$ cm $= 0.09$ m
Charge $Q = 3$ nC $= 3 \times 10^{- 9}$ C
Potential at a distance 9 cm, $V = \frac{k Q}{r}$ where $k = 9 \times 10^{9}$
$∴$ $V = \frac{9 \times 10^{9} \times 3 \times 10^{- 9}}{0.09} = 300$ V