Question

what is the difference in potential between two points one at 10cm and other at 20cm distance from a charge of -5.5microcoloumb

Answer 1

Answer:

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Answer 2

Answer:

The potential difference between the two points is -2.475×10⁵volt.

Explanation:

The difference in potential between the two points is given as,

$V=kQ(\frac{1}{r_1}-\frac{1}{r_2})$       (1)

Where,

V=potential difference

k=coulomb's constant=9×10⁹Nm²C⁻²

Q=magnitude of charge on the point charge

r₁,r₂=respective distance at which potential difference is to be found

From the question we have,

Q=-5.5×10⁻⁶C

r₁=10cm=0.1m

r₂=20cm=0.2m

By substituting all the required values in equation (1) we get;

$V=9\times 10^9\times -5.5\times 10^-^6(\frac{1}{0.1}-\frac{1}{0.2})$

$V=-49.5\times 10^3\times \frac{1}{0.2}$

$V=-247.5\times 10^3$

$V=-2.475\times 10^5volt$

Hence, the potential difference between the two points is -2.475×10⁵volt.