Question

Two equal charges of 2×10^-16 C are kept 20cm apart in air. At point midway between them find electric field intensity and force acting on the charge 2×10^-16 C ???

Answer 1

hre is your answer buddy

Answer 2

Concept:

Coulomb's law can be defined as

F = kq₁q₂/r²

where q₁, q₂ are the charges, and r is the distance between them.

Given:

Two equal charges, 2×10⁻¹⁶ C

Distance between the charges, d = 20 cm = 0.2 m

Find:

Electric field intensity and force acting on the charge 2×10⁻¹⁶ C at a point midway between the charges.

Solution:

Electric field intensity at midway is zero, as both the charges are the same, and that is why the electric field is canceled by them.

Force on each charge due to other charges by Coulomb's law,

F = kq₁q₂/r²

F = (9×10⁹×2×10⁻¹⁶ ×2×10⁻¹⁶ )/(0.2)²

F = 9×10⁻²¹ N

Hence, the electric field intensity is zero and the force acting on the charge is  9×10⁻²¹ N.

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