Question

two points A and B are located at distance of 100cm from an electric dipole. A is an axial point while B is in perpendicular bisector. The electric field intensity at is vector E. Then the intensity at B will be​

Answer 1

The intensity at point B will be EB = 1/2EA.

Explanation:

Two points A and B are located at distance =100 cm = 100 x 10^-2 m

The electric field intensity at A= E  

Then, the intensity at B will be.

E1/F EA / EB = 100 / 50

EB = 1/2EA

Thus the intensity at point B will be EB = 1/2EA.

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The electric potential on the surface of a sphere of radius R and charge 3 x 10^-6 C is 500 V then find the electric field intensity on the surface of the sphere​ ?

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