Explanation

• An equipotential surface has the same electric potential everywhere.
• No potential difference exists between any two points on it.
• Work done depends on potential difference.
• If potential difference is zero, work done is also zero.
• So moving charge on it requires no energy.

Last verified on 22-05-2026

Explanation

• A conductor contains free electrons that can move freely.
• In electrostatic equilibrium, charges stop moving.
• If any electric field existed inside, charges would move.
• So charges rearrange themselves until internal field cancels out.
• Hence, no net electric field remains inside.

Last verified on 22-05-2026

Explanation

When lenses are placed in contact, their powers add:

P = P₁ + P₂ = +2 D + (-5 D) = -3 D

Focal length f in meters is f = 1/P:

f = 1/(-3) = -0.33 m

So the equivalent power is -3 D and the equivalent focal length is -0.33 m.

Last verified on 22-05-2026

Explanation

For a black body, radiated energy per unit time per unit area follows Stefan-Boltzmann law:

E ∝ T⁴

Temperature goes from 400 K to 800 K, so it doubles: 800/400 = 2.

New energy = (2)⁴ × original energy = 16 × original energy.

Last verified on 22-05-2026

Explanation

For light passing from a medium to air, the critical angle is given by:

sin θc = 1/n

Here n = 1.5, so  

sin θc = 1/1.5 = 0.6667  

θc = arcsin(0.6667) ≈ 41.8°

Last verified on 22-05-2026

Explanation

For masses m and 3m over a frictionless pulley, the heavier mass pulls the system down. 

Net pulling force = 3mg - mg = 2mg  

Total mass being moved = m + 3m = 4m  

So acceleration a = net force / total mass = 2mg / 4m = g/2.

Thus the system accelerates at g/2, with 3m going down and m going up.

Last verified on 22-05-2026

Explanation

Find acceleration first using F = ma

a = F/m = 6 N / 2 kg = 3 m/s²

Initial velocity u = 0 since it starts from rest.  

Time t = 3 s.

Use the distance formula: s = ut + ½at² 

s = 0(3) + ½(3)(3)²  

s = ½(3)(9)  

s = 13.5 m

Last verified on 22-05-2026

Explanation

At the midpoint:

• Field due to +q (at −a) points away from +q → toward +x direction
• Field due to −q (at +a) points toward −q → also toward +x direction

Both contributions add in the +x direction toward the −q charge, so net field is along +x axis.

So correct direction is actually +x axis.

Last verified on 22-05-2026

Explanation

First, series combination:

1/C = 1/2 + 1/3 = (3+2)/6 = 5/6

C = 6/5 = 1.2 μF

Now in parallel with 5 μF:

C_total = 1.2 + 5 = 6.2 μF

Last verified on 22-05-2026

Explanation

• Electric potential due to a point charge is: V=kQ/r.
• It is inversely proportional to distance r.
• k = Coulomb constant (9 × 10⁹ Nm²/C²).

Last verified on 22-05-2026