Test: Electrostatic Potential and Capacitance - 2 - CUET Humanities MCQ

The work done in carrying a charge q around a circle of radius r with a charge Q at the centre is zero as work done over equipotential surface is zero.
Around the point charge, equipotential surfaces are the concentric spheres of different radii.

The distance of point P₁ from charge +q is r₁ = z - a and from charge -q is r₂ = z + a.

The capacitance (C) of a capacitor is a measure of the amount of charge (Q) stored on each plate for a given potential difference (V).
∴ C = Q/V farad
The capacitance depends upon the size of the conductor, and not upon the nature, thickness, or colour of the conductor.
For example, capacitance of the parallel plate capacitor is , where A is the area of the plates and 'd' is the separation between the plates.

Electric field E = -
Force on charge (-q) = -qE = +8qx
At x = 0.5 m, force = 8 x 2 x 10^-6 x 0.5 = 8 x 10^-6 N
Hence, the correct option is (4).

Since potential at P due to the charge q is ,
Potential at Q due to the charge q is 
Since OQ > OP, so V_Q < V_P.
Potential energy of the electron at P:

Potential energy of the electron at Q:

U_Q is less negative than U_P.
Hence, potential energy of the electron is more at Q than that at P.

Given, C₂ = 3C₁
Hence, 
This gives  which is option (2).

Charge on capacitor plates without the dielectric is Q = CV = (5 x 10^-6 F) x 1V = 5 x 10^-6C = 5μC 
The capacitance after the dielectric is introduced is 

∴ Charge on capacitor plates now will be 

Additional charge transferred = Q' - Q = 10μC - 5μ​​​​​​​C = 5μC.

Energy stored initially is U_i = Q²/2C. If d is doubled, C becomes C/2. Hence, energy stored when d is doubled is Uf = Q²/C
∴ Work needed is 

Now C  Hence W =  which is choice (3). 

The batteries are in opposition as their positive terminals are connected together. Hence, the effective voltage is:
V = V₁ – V₂ = 12 – 2 = 10 V
As the capacitors C₁ and C₂ are in series, the effective capacitance of the circuit is given by:

Or C = 6/5 = 1.2μF 
Therefore, charge on capacitors, Q = CV = 1.2 × μF 10V = 12 μC
∴ Potential difference across A and B, i.e. across capacitor C₂  = Q/C₂ = 12μC​​​​​​​/2μF = 6 V

By energy conservation,
U_series = U_parallel
1/2(C/n)v² = 1/2(nC)v'²
v/n = v'
So, energy remains same and potential difference is v/n.
Hence, (3) is the correct answer.

When S₁ is closed, charge on plate 2 will not change because this plate is isolated.
The bounded charges will not change, so the potential difference across the capacitors will remain the same.
When S₃ is closed, plates 1 and 4 of C₁ and C₂ remain isolated.
So, the bounded charges remain intact.
Therefore, potentials V₁ and V₂ will remain the same.
When S₁ and S₂ are closed, charges on plates 2 and 3 remain the same. So, the bounded charges on plates 1 and 4 do not change.
So, V₁ and V₂ will not change.
When switches S₁ and S₃ are closed, charges on plates will remain the same.
So, the bounded charges will not change.
So, V₁ and V₂ will remain the same.

Charge on electron (–e) = –1.6 × 10^–19 C
Charge on proton (e) = 1.6 × 10^–19 C
Separation r = 0.53 Å = 0.53 × 10^–10 m
If the zero of potential energy is taken to be at infinite separation, then the potential energy of the electron-proton system is:

Hence, the correct option is (2).

Potential of the bubble is V = KQ/R
where K is constant, Q is charge and R is radius.
When radius becomes nr(Q remains same)
New potential V' = KQ/nR
Thus V' = V/n 
Hence, the correct option is (3).

The series combination of C₂ and C₃ is equivalent to a capacitance C'.
1/C' = 1/C₂ + 1/C₃
Or 

C'' = 100 pF + 100 pF = 200 pF
The equivalent capacitance between A and D is:
1/C = (1/200) + (1/100)
C = 200/3 pF

The distance of a vertex from the centre of the cube of side b is r = √3 b/2
The potential due to charge q (located at vertices) at the centre is q/4 
Hence, the potential due to the arrangement of eight charges (each of magnitude q) at the centre is