Question

Two charges 5 microcoulomb,-3 microcoulomb are separated by a distance of 40cm in air. Find the location of a point on the line joining the two charges where the electric field is zero

Answer 1

Two charges 5 microcoulomb,-3 microcoulomb are separated by a distance of 40cm in air. The location of a point on the line joining the two charges where the electric field is zero is 1 m.

• Given,
• Charge qA at point A= 5 microcoulomb
• Charge qB at point B= -3 microcoulomb
• Distance between the charges d = 40 cm = 0.4 m
• Let, the point where the potential is 0  = O
• Assume, the distance of O from point A  = x
• and, the distance of O from point B  = 0.4 - x
• Since the electric potential is zero at the point O, then the potential at points A and B are equal.
• ∴ vA = vB
• $\dfrac{1}{4\pi e_0} * \dfrac{5*10^{-6}}{x} = \dfrac{1}{4\pi e_0} * \dfrac{-3*10^{-6}}{0.4 - x}$
• $\dfrac{5}{x} = \dfrac{-3}{0.4-x}$
• $2 - 5x = -3x\\\\2 = 2x\\\\x = 1$
• ∴ x = 1 m

Answer 2

Answer:

At point o the net electric field is zero at the distance is 0.25 m

Explanation:

Given as ;

Two charge $q_1$ = 5 $\mu$c  , $q_2$ = - 3 $\mu$c

The distance between the charges = d = 40 cm = 0.4 m

Let The point at which electric field between them zero = o

Since, electric potential at o is zero

So, potential at o due to $q_1$  = potential at o due to $q_2$

Or, $V_1 + V_2$  = 0

i.e $V_1 = - V_2$

Or, $\dfrac{q_1}{4\Pi \varepsilon _0x}$   = - $\dfrac{q_2}{4\Pi \varepsilon _0(0.4- x)}$

Or, $\dfrac{5\mu c}{4\Pi \varepsilon _0 x}$  = - $\dfrac{- 3\mu c}{4\Pi \varepsilon _0 (0.4 - x)}$

Or, $\dfrac{5}{x}$ = $\dfrac{3}{0.4-x}$

or, 5 (0.4 - x) = 3 x

or, 2 - 5 x = 3 x

Or, 8 x = 2

∴   x = $\dfrac{2}{8}$

i.e x = 0.25 m

So, At point o the net electric field is zero at the distance = x = 0.25 m

Hence,  At point o the net electric field is zero at the distance is 0.25 m Answer