Question

39
17. Two charges of +2.4uc and -2.4uc are at a distance
2.5 x 10-3 m apart from each other. Determine the
magnitude of the electric field in the broad-side-on
position at a distance of 0.30 m from this dipole. If the
dipole be rotated through 90°, then what will be the
intensity of the field ?
Ans 20 X 193 NC-1, 4.0 x 10° NC1.​

Answer 1

Answer:

20x193 NC

Explanation:

this is the answer

Answer 2

Answer:

We know  

Electric filed on broad side on position is given by

E=2qak/r^3

(r>>a)

Given that r = 0.3 m

2a = 2.5 × 10–3 m

q = 2.4 × 10–6 C

So,

e=(2.4 *10^-6 )*(2.5 *10^-3)*(9* 10^9)/(0.3)^3

=20 X 193 NC-1

As we rotate the dipole by 90°  

Intensity of the field will not be affected.  That will be same.

hope this is helpful