Question

The electric field intensity at a point
situated 4 metres from a point charge
is 200 N/C. If the distance is reduced to
2 metres, the field intensity will be
(a) 400 N/C (6) 600 N/C
(c) 800 N/C (d) 1200 N/C

Answer 1

Answer:

correct option is (c) 800N/C

Explanation:

see...............

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Answer 2

The field intensity will be 800 N/C.

Explanation:

The electric field intensity is given by

$E=k\frac{Q}{d^2}$

Here, Q is the point charge k is the proportionality constant and d is the distance between the charges.

Now E at a point  situated 4 m from a point charge,

$E_1=k\frac{Q}{d_1^2}$

Similarly, if the distance is reduced to  2 m, then

$E_2=k\frac{Q}{d_2^2}$

Therefore,

$\frac{E_1}{E_2} =\frac{k\frac{Q}{d_1^2}}{k\frac{Q}{d_2^2}}$

$\frac{E_1}{E_2} =\frac{d^2_2}{d^2_1}$

Substitute the given values, we get

$\frac{200N/C}{E_2}=\frac{(2)^2}{(4)^2}$

$E_2=800N/C$.

Thus, the field intensity will be 800 N/C.

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