Question

24. Calculate the intensity of electric field at a point 25 cm from a charge of 4.8uC
in a medium of dielectric constant 3.6.
(a) 19.2x10'NC
(b) 192X10‘N/C
(c) 192X10²N/C
(d) 1.92X10’N/C​

Answer 1

Answer:

option b

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Answer 2

Dear Student,

● Answer -

1.92×10^5 N/C

◆ Explaination -

# Given -

q = 4.8 uC = 4.8×10^-6 C

x = 25 cm = 25×10^-2 m

k = 3.6

# Solution -

Intensity of electric field at a point x cm away from charge is -

E = (1/4πε0k) × q / x^2

E = (9×10^9/3.6) × 4.8×10^-6 / (25×10^-2)^2

E = 12000 / 0.25^2

E = 192000 N/C

Hence, intensity of electric field is 1.92×10^5 N/C.

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