Question

22. Two positive charges of 1 uC and 2 uC are placed
1 metre apart. The value of electric field in N/C
at the mid point of the two charges will be :-
(1) 10.8 x 104
(2) 3.6 x 104
(3) 1.8 x 104
(4) 5.4 x 104​

Answer 1

Answer:

2

Explanation:

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E1 = kq/d²

= kq/(1/2)²

= 4kq

E2 = kQ/d²

= 4kQ

Net electric field is = E2 - E1

= 4k(Q-q)

= 4k(2-1)

= 4*9*(10^9)(10^-6)

= 3.6× 10^4

Answer 2

Given that :

Charge 1, $q_1=1\ \mu C$

Charge 2, $q_2=2\ \mu C$

Distance between charges, d = 1 m

To find :

The value of electric field in N/C  at the mid point of the two charges.

Solution :

The electric field due to charge 1 is :

$E_1=\dfrac{kq}{r^2}\\\\E_1=4kq_1$

The electric field due to charge 2 is :

$E_2=\dfrac{kq_2}{r^2}\\\\E_1=4kq_2$

Net electric field at the mid point of the two charges is :

$E=E_2-E_1\\\\E=4kq_2-4kq_1\\\\E=4k(q_2-q_1)\\\\E=4k(2-1)\times 10^{-6}\\\\E=4\times 9\times 10^9\times 10^{-6}\\\\E=3.6\times 10^4\ N/C$

So, the correct option is (2).

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