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Solutions
Concept:
Electric Field Intensity: The electric field intensity at any point is the strength of the electric field at the point.
It is defined as the force experienced by the unit positive charge placed at that point.
\(\vec E = \frac{{\vec F}}{{{q_o}}}\)
Where F = force and qo = small test charge
The magnitude of the electric field (E) is given by:
\(E = \frac{{kq}}{{{r^2}}}\)
Where K = constant called electrostatic force constant
q = source charge
r = distance
Calculation:
Given that: Distance between two charges (r) = 8 cm
Charges on each (q) = + 5 μC

Let P be the point at distance x from A, where the net electric field is zero.
The electric field at P due to A is
\({E_1} = \frac{{k\; \times \;5\; \times \;{{10}^{ - 6}}}}{{{x^2}}}\)
The electric field at P due to B is
\({E_2} = \frac{{k\; \times\; \left( { 5\; \times \;{{10}^{ - 6}}} \right)}}{{{{\left( {8\; - \;x} \right)}^2}}}\)
At point P, E1 - E2 = 0
E1 = E2
\( \frac{{k\; \times \;5\; \times \;{{10}^{ - 6}}}}{{{x^2}}} = \frac{{k\; \times \;5\; \times \;{{10}^{ - 6}}}}{{{{\left( {8\; -\; x} \right)}^2}}}\)
\( \frac{1}{{{x^2}}} = \frac{1}{{{{\left( {8\; - \; x} \right)}^2}}}\)
By square rooting both side, we get:
x = 8 - x
2x = 8
x = 4 cm