The electrical potential on the surface of a sphere of radius r due to a charge 3× 10-6C is 500 V. The intensity of electric field on the surface of the sphere is [ (1/4π ε 0)=9× 109Nm2C-2 ] (in text NC-1) :

Question

The electrical potential on the surface of a sphere of radius r due to a charge 3× 10-6C is 500 V. The intensity of electric field on the surface of the sphere is [ (1/4π ε 0)=9× 109Nm2C-2 ] (in text NC-1) : (A) 250/27 (B) 27/250 (C) 250 (D) 27

Tardigrade Admin · Accepted Answer

given V=500 v beginarrayl ⇒ ( kq / r )=500 volts ⇒ r =(9 × 109 × 3 × 10-6/500)=(27 × 1000/500)=54 endarray we know E=(k q/r2) ⇒ E =(9 × 109 × 3 × 10-6/54 × 54) ⇒ E =(27 × 250/27) ⇒ E =(250/27)

Q. The electrical potential on the surface of a sphere of radius r due to a charge 3×10−6C is 500 V. The intensity of electric field on the surface of the sphere is [4πε0​1​=9×109Nm2C−2] (inNC−1) :

250/27

27/250

250

27

given V=500v <br/><br/>⇒rkq​=500volts<br/>⇒r=5009×109×3×10−6​=50027×1000​=54<br/>​<br/> we know E=r2kq​ <br/>⇒E=54×549×109×3×10−6​<br/> <br/>⇒E=2727×250​<br/> <br/>⇒E=27250​<br/>

Q. The electrical potential on the surface of a sphere of radius $r$ due to a charge $3 \times 10^{- 6} C$ is 500 V. The intensity of electric field on the surface of the sphere is $[\frac{1}{4 π ε _{0}} = 9 \times 10^{9} N m^{2} C^{- 2}]$ $(in N C^{- 1})$ :