Question

A point P, is 40 m away from charge 2 uC and 20 m away from charge 4 uС. Find the
electric potential at point P. How much work to be done for a charge of -0.4 C brought
from infinity to point P ?
​

Answer 1

Answer:

Answer is=

The test change is an electron

∴ Work done W=qΔVwhere,

q=−16×10

−19

C,V

A

=20V

V

B

=−40V

∴W=(−16×10

19

)[−40−20]

=9.6×10

−18

J

Answer 2

Answer:

Potential at point P is 2250 Volts.

The work needed to be done is 900 J.

Explanation:

Given,

q₁ = 2 μC, q₂ = 4μC

And the distance from the point P to the charges q₁ and q₂ are r₁ and r₂ respectively.

To find,

Electric potential at point P (V)

We know that electric potential for charge q₁ is

$V_1 = \frac{kq_1}{r_1}$

$\implies V_1 = \frac{9\times 10^9\times 2\mu C}{40 m} \\\implies V_1 = \frac{9\times 10^9\times 2\times 10^{-6} C}{40 m}\\\implies V_1 = 450 Volts$

We also know that electric potential for charge q₂ is:

$V_2 = \frac{kq_2}{r_2}$

$\implies V_2 = \frac{9\times 10^9\times 4\mu C}{20 m} \\\implies V_2 = \frac{9\times 10^9\times 4\times 10^{-6} C}{20 m}\\\implies V_2 = 1800 Volts$

Therefore, potential at point P is V = V₁ + V₂ = 450 + 1800 =2250 Volts.

Now for the finding of the work to be done(W) for a charge of -0.4 C brought from infinity to point P is

W = qV

W = -0.4(2250)

W = 900 J

Therefore, the work needed to be done is 900 J.

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