Question

The potential at point to charge of 5×10-7 c located 10 CM away is

Answer 1

Answer:

Given:

Charge = 5 × 10⁻⁷ C

The distance between the charge and the point = 10 cm = \frac{10}{100}

100

10

m = 0.1 m

To find:

Electric Potential at the point

Solution:

We know that at any point, the electric potential at a point due to a charge separated by a distance is given by,

\boxed{\bold{V = k \frac{q}{r} }}

V=k

r

q

where

V = electric potential

k = Coulomb's constant = 9 × 10⁹ N⋅m²/C²

q = charge

r = distance of separation

Now, we will substitute the given values of q and r in the formula above,

V = 9 × 10⁹ ×

⇒ V = 450 × 10⁹ × 10⁻⁷

⇒ V = 450 × 10⁹⁻⁷

⇒ V = 450 × 10²

⇒ V = 45000 ×

⇒ V = 4.5 × 10⁴ volts

Thus, the electric potential at a point due to charge of 5 × 10⁻⁷ C located 10 cm away is 4.5 × 10⁴ volts.

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Also View:

Find the electric potential at a point if 0.002joule work is done to bring a 40mc charge from infinity

THE ELECTRIC FIELD DUE TO A POINT CHARGE AT A DISTANCE 6m FROM IT IS 630N/C. FIND THE MAGNITUDE OF CHARGE.

Answer 2

Answer:

45×10²volt

Explanation:

v=KQ/r

v=9×10⁹×5×10-⁷/0.1

v=45×10²volt