Physics, asked by kumarshiv6031, 11 months ago

Two point charges q1 = 400 micro coulomb and q2=100 micro coulomb are kept fixed 60 cm apart in vacuum find intensity of the electric field at mid point of the line joining q1 and q2

Answers

Answered by Anonymous
7

Given - q1 = 400 micro coulomb q2=100 micro coulomb

Distance - 60 cm

Find - Intensity of electric field at mid point of the two charges.

Solution - The formula to calculate intensity of electric field is as follows-

E = (1/4π€)(Q/r²)

In the formula, E refers to electric field, (1/4π€) can be written as 9*10^9, Q refers to charges. R² denotes the point at which Intensity of elctric field is to be calculated.

Since the intensity is to be calculated at mid point, the equation will be -

E = E1 - E2

R will be 30 cm(due to mid point) = 0.3 m.

Q1 for E1 - 400 microC = 4*10^-4 C

Q2 for E2 - 100 microC = 1*10^-4 C

E1 = 9*10^9*4*10^-4/(0.3)²

E1 = 4*10^7 NC

E2 = 9*10^9*1*10^-4/(0.3)²

E2 = 1*10^7 NC

Therefore, E = 4*10^7 - 1*10^7

E = 3*10^7 NC

Hence, intensity of electric field at mid point will be 3*10^7 NC.

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