Question

calculate the potential at a point 20cm away from the point charge 1C​

Answer 1

Answer:

Potential 'V' due to a point charge 'q' is

$V = \frac{kq}{R}$

k = 9*10⁹ Nm²/C²

q = 1C

R = 0.2m

V = (9*10⁹ Nm²/C² x 1C)/ 0.2m = 4.5 x 10¹⁰ V

Answer 2

Answer:

Potential at point 20 centimeters of a charge 1 c is 45×10⁹.

Explanation:

here, d =20 cm implies 0.2 m and charge c=1 C

Potential of distance d 20 cm,      

$potential = \frac{k×c}{d}$

 

where k=9×10⁹

$potential= \frac{9×10⁹×1}{0.2} = 45×10⁹ \: v$

therefore Potential at distance if 20 cm for 1 c charge is 45×10⁹.

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