Question

the electric intensity at a point 50 cm from a charge of 3.2 micro coloumb ina medium of di electric constant 2 is equal to ​

Answer 1

Dear Student,

● Answer -

E = 57600 N/C

◆ Explaination -

# Given -

q = 3.2 uC = 3.2×10^-6 C

r = 50 cm = 0.5 m

k = 2

# Solution -

Electric intensity at a point is calculated by -

E = (1/4πε0k) q/r²

E = (9×10^9/2) × 3.2×10^-6 / 0.5²

E = 57600 N/C

Hence, electric intensity at given point is 57600 N/C.

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