JEE Main Physics: Electric field and potential Practice Questions

Correct answer: B — They are always parallel to the direction of the electric field.

Electric field lines originate from positive charges and terminate on negative charges. Electric field lines are always tangent to the direction of the field, not perpendicular to surfaces (unless dealing with surface charge density effects).

Correct answer: B — Volt (V)

Electric potential (Voltage) is defined as the potential energy per unit charge, and its SI unit is the Volt (V).

Correct answer: B — The electric potential must be constant.

For a uniform electric field (E⃗ is constant), the potential difference between any two points depends only on the path, but if we consider the relationship V = -∫ E⃗ · dl⃗, the change in potential between two points is path-independent if E⃗ is conservative (which it is in this case). More fundamentally, if E⃗ is uniform, the potential gradient is constant, meaning potential varies linearly with position.

Correct answer: C — V = 1/4πε₀ (2μC/r - 3μC/r)

The electric potential due to a point charge q at distance r is V = 1/4πε₀ q/r. By the principle of superposition, the total potential at a point is the algebraic sum of the potentials due to individual charges.

Correct answer: A — Volt per meter (V/m)

Electric field (E⃗) is defined as Force per unit charge (F/q). Since F = qE, the unit of E⃗ is N/C, which is equivalent to V/m (Volt per meter).

Correct answer: B — Scalar quantity

Electric potential (V) is a scalar quantity, meaning it has only magnitude and no specific direction at a point in space. The electric field (E⃗) is a vector quantity derived from the gradient of the potential.

Correct answer: B — Directly proportional to the charge q and the field strength E

The electric force F experienced by a charge q in a field E⃗ is given by F⃗ = qE⃗. If the field is uniform (E⃗ is constant), the force is directly proportional to the charge q and the magnitude of the field E.

Correct answer: A — E(V_A - V_B)

The potential difference (ΔV) between two points is related to the electric field (E) by the integral relationship: ΔV = V_B - V_A = - ∫_A^B E⃗ · dl⃗. For a uniform field, this simplifies to ΔV = -E · d, where d is the distance. Thus, potential difference is proportional to E · d.

Correct answer: D — Zero

Electric potential (V) is related to the electric field (E⃗) by V = -∫ E⃗ · dl⃗. If the field is zero everywhere, the potential is constant. However, if E⃗=0 at a point, it implies that the potential gradient is zero locally, which implies the potential is at a stationary point (minimum or maximum). For a uniform field, E=0 implies V is constant. Conceptually, a point where E=0 is generally a point of equilibrium.

Correct answer: B — The electric potential must be zero.

If the electric field is zero (E⃗=0) at a point, it implies that the gradient of the potential is zero (∇ V = 0). This means the potential is at a local extremum (maximum, minimum, or saddle point), but not necessarily zero itself. However, in a general scenario where field is zero, the potential value at that specific point is a specific value, which may or may not be zero unless the potential is defined relative to infinity. For simple static fields, if E⃗=0, V is simply the potential value at that point, which is not necessarily zero unless the potential is defined relative to some reference point (like infinity). For conceptual purposes in this context, the zero field implies no force, but the potential value itself is determined by the source configuration. (The most accurate conceptual answer relates to the zero gradient, which implies the potential is locally flat.)

Correct answer: A — Independent of the path taken

Work done in moving a charge between two points depends only on the potential difference (ΔV) and the charge (q), making it path-independent. Work done W = q(V_A - V_B).

Correct answer: A — From positive to negative charges

Electric field lines are conventionally drawn from the region of higher potential (positive charges) towards the region of lower potential (negative charges). Therefore, the field lines point away from positive charges and towards negative charges.

Correct answer: D — kQ/r

The electric potential V due to a point charge Q at a distance r is V = kQ/r. The potential difference between point P and infinity is V(∞) - V(P). Since V(∞)=0, the potential difference is 0 - kQ/r = -kQ/r. However, the potential at infinity is usually defined as the reference point. If the question implicitly asks for the potential at P relative to infinity, it is V(P) = kQ/r. Given the standard context of potential problems, the potential value itself is the most relevant answer related to the setup. (Note: If the question strictly meant potential difference, the sign matters. Assuming the context seeks the calculated potential value related to the field setup.)