Answer :

43 mV

Given,

Mass of the particle, m10 mg

The charge of the particle, q -0.01 μC -0.01 10^{-6} C

The capacitance of each pair of the parallel capacitor plates, C0.04 μF 0.0410^{-6} F

Area of each capacitor plates, A 100 cm^{2}10010^{-4} m^{2}

V is the potential difference required for the particle to be in equilibrium?

**Formula used**

For the particle of mass ‘m’ to stay in equilibrium in the given set up, the weight of the particle W) should be opposed by the electric force F), acting on the same charged particle. The electric force is exerted by the electric field in between the capacitor plates. As the weight is acting downward, the electrical force should act upward for the equilibrium.

So,

Or

Where,

g Acceleration due to gravity 9.81m/s^{2}

q charge of the particle -0.01 μC -0.01 10^{-6} C;

m10 mg10×10^{-4}kg;

E Magnitude of Electric field in between the capacitor plates;

But from Gauss’s law, we have,

Where,

Q Charge on the capacitor plates same on both capacitors for series arrangement)

ε_{0}Permittivity of free space 8.85 10^{-12}Fm^{-1}

A= Area of the plate in the parallel plate capacitor10010^{-4} m^{2}

We know that, for capacitors connected in series across the voltage V, the effective capacitance, C_{eff} will be

Or,

Here C_{1}=C_{2}= C = 0.04 μF

Hence,

Or,

With this, we can calculate the value of charge stored Q) in the given capacitor arrangement as,

Where, V is the potential difference required to produce enough electric field to oppose the weight of the particle.

Putting eqn.3 in eqn.2, we get,

Now, substituting eeqn.4 in eqn.1, we get,

Or,

Substituting the known values, we get

Or,

V=0.434V=43.4 mV

Hence, to keep the particle of mass 10mg, the potential difference in the set up should be 43 mV.

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