Question

the potential at a certain distance from a point charge is 600volt and the electric field is 200NC*-1. Find 1. The distance of the point from the charge. 2. magnitude of the charge.​

Answer 1

Answer:

Given Data

The potential at certain distance from a point charge is

V

=

600

V

.

The electric field due to the point charge is

E

=

200

N

/

C

.

a) The expression of the potential due to the point charge is given as,

V

=

k

q

r

The expression of the electric field due to the point charge is given as,

E

=

k

q

r

2

Dividing both the equation as,

V

E

=

r

r

=

600

200

=

3

b) The point charge is calculated as,

V

=

k

q

r

600

=

9

×

10

9

×

q

3

q

=

3

×

600

9

×

10

9

=

2

×

10

−

7

C

Answer 2

The distance of the point from the charge is 3m.

The magnitude of the point charge is  2 * $10^{-7}$ C

Given:

Potential V = 600 volts and Electric field = 200 N/C.

To Find:

1) The distance of the point from the charge

2) Magnitude of the point charge

Solution:

1) V = kq / r

⇒ 600 = kq / r --------------(1)

E = kq /r²

⇒ 200 =  kq /r² --------------(2)

Dividing Equation (1) and (2)

600 / 200  =   (kq / r) /(kq /r²)

⇒ r = 3

Therefore, The distance of the point from the charge is 3m.

2) V = kq/r

⇒ 600 = (9 * $10^{9}$ * q ) / 3

⇒ q = 600 / (3 * $10^{9}$ )

⇒ q =  2 * $10^{-7}$ C.

The magnitude of the point charge is  2 * $10^{-7}$ C

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