The potential at point to charge of 5×10-7 c located 10 CM away is
Answers
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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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:
45×10²volt
Explanation:
v=KQ/r
v=9×10⁹×5×10-⁷/0.1
v=45×10²volt