Electrostatic Potential - Free MCQ Practice Test with solutions, NEET Physics

The standard convention states that the electric potential at a point infinitely far from a charge is considered to be:

• Zero

This means that as you move further away from a charge, the potential energy associated with that charge diminishes to zero.

Solution:

To calculate the work done in moving a charge in an electric field, we use the formula:

• Work = Charge × Voltage Difference

Here, we have:

• Charge = -2 x 10^-9 C
• Initial position P = (0, 3 cm, 0)
• Final position Q = (0, 4 cm, 0)

The voltage (potential) at a distance r from a point charge is given by:

• V = k × Q / r
• Where k = 9 x 10⁹ Nm²/C² and Q = 6 mC = 6 x 10^-3 C

Calculating the voltage at points P and Q:

• At P (0, 3 cm): r = 0.03 m
• V_P = (9 x 10⁹) × (6 x 10^-3) / 0.03 = 1.8 x 10⁶ V
• At Q (0, 4 cm): r = 0.04 m
• V_Q = (9 x 10⁹) × (6 x 10^-3) / 0.04 = 1.35 x 10⁶ V

Now, the voltage difference:

• ΔV = V_Q - V_P = 1.35 x 10⁶ - 1.8 x 10⁶ = -0.45 x 10⁶ V

Finally, calculating the work done:

• Work = (-2 x 10^-9) × (-0.45 x 10⁶) = 0.0009 J = 0.9 J

Thus, the work done is 0.9 J.

Potential difference between two points is given by

V_a - V_b = W/q₀

Work, W = 2 J

Charge, q₀ = 20 C

Potential difference = 2/20 = 0.1 V

The correct option is C.

Electric field at a distance r,
(r₀ < r < 2r₀) from the center is given by

Potential at the outer surface,

∴ If V is the required potential at r  = 3r₀/2,

E.l = V   where,
E = electric field intensity = 12 N/C
V = electric potential = 6 J/C
=> distance between A and B, 
l = (6 / 12) m or (1 / 2) m = 0.5 m
 

The overall work performed in carrying a 2coulomb charge in a circular orbit of radius 3 m around a charge of 10 coulomb is calculated below.
It is a well-known fact that W=qdv.
Here dV is the change in overall potential. In the circular orbit of r potential at each point is similar.
Most significantly, the value of r is 3.  
The value of dv=0 and hence W=q₀=0.

The electric potential difference (V) is defined as the work done (W) per unit charge (q):

V = W / q

• Work done (W) has the dimensional formula: [M L² T⁻²]

• Charge (q) has the dimensional formula: [A T]

Now divide the two:

[V] = [M L² T⁻²] / [A T] = [M L² T⁻³ A⁻¹]

Hence, the dimensional formula of potential difference is: [M L² T⁻³ A⁻¹]

From the definition, the work done to a test charge ‘q0’ from one place to another place in an electric field is given by the formula
W=q₀x[v_final-v_initial ]
100=4x[v-(-5)]
v+5=25
v=20V
 

The absolute electric potential at the origin due to two point charges can be calculated using the formula:

• V = k (q₁/r₁ + q₂/r₂)

Where:

• V = electric potential
• k = 8.99 x 10⁹ N m²/C² (Coulomb's constant)
• q₁ = 3 µC (charge at x₁ = 6 cm)
• q₂ = 2 µC (charge at x₂ = -4 cm)
• r₁ = 6 cm = 0.06 m (distance from origin to q₁)
• r₂ = 4 cm = 0.04 m (distance from origin to q₂)

Substituting the values:

• V = 8.99 x 10⁹ × (3 x 10^-6/0.06 + 2 x 10^-6/0.04)

Calculating each term:

• For q₁: 3 x 10^-6/0.06 = 5 x 10^-5
• For q₂: 2 x 10^-6/0.04 = 5 x 10^-5

Therefore:

• V = 8.99 x 10⁹ × (5 x 10^-5 + 5 x 10^-5)
• V = 8.99 x 10⁹ × 10 x 10^-5 = 8.99 x 10⁵ V

Thus, the absolute electric potential at the origin is: 9 x 10⁵ V.

There will be no loss of energy if the potential of the spheres is the same i.e. if

Electric potential is the electric potential energy per unit charge. In equation form, V=U/q, where U is the potential energy, q is the charge, and V is the electric potential. Since both the potential energy and charge are scalar quantities, so does the potential.
And it's dimension [ L²M¹T^-3I^-1]

Let λ be the charge density on the inner cylinder.

For a < r < b, we get

Hence, a potential difference appears between the two cylinders when a charge density is given to the inner cylinder.

When equal charges are given to two spheres of different radii, the potential will be more or the smaller sphere as per the equation, Potential = Charge / Radius.
Since potential is inversely proportional to radius, the smaller radius will have higher potential and vice versa.

By dimensional analysis

but by superposition

Because of the centre of the larger cube lies at a corner of the eight smaller cubes of which it is made
therefore,

Assuming ring as dipole, then dipole moment   and  are in same direction, so potential energy U =−PE