Questions on electric potential

Question 1

Define the absolute electric potential at a point in an electric field.

Accepted Answer

Work done per unit positive charge (1); in bringing a small test charge from infinity to the point (where the potential at infinity is zero) (1).

Question 2

Calculate the electric potential at a distance of 0.10 m from a point charge of 5.0 × 10⁻⁹ C. k = 8.99 × 10⁹ N m² C⁻²

Accepted Answer

V = kQ/r = (8.99 × 10⁹ × 5.0 × 10⁻⁹)/0.10 (1) V = 450 V (1)

Question 3

Describe the equipotential surfaces around an isolated point charge and state how they are related to the electric field lines.

Accepted Answer

Concentric spheres centred on the charge (1); everywhere perpendicular to the radial field lines (and no work is done moving a charge along an equipotential) (1).

Question 4

A charge of 5.0 μC is moved 0.40 m along an equipotential surface. State the work done on the charge.

Accepted Answer

Zero: the potential difference along an equipotential is zero (1).

Question 5

The electric potential at a distance r from a point charge is 800 V. State the potential at a distance 2r, and at a distance 4r, from the charge.

Accepted Answer

V ∝ 1/r, so at 2r the potential is 400 V (1) and at 4r it is 200 V (1).

Question 6

State what a potential difference of one volt means, in terms of energy and charge.

Accepted Answer

One joule of energy is transferred per coulomb of charge moved through it (1).

Question 7

Calculate the electric potential at a distance of 0.20 m from a point charge of −3.0 × 10⁻⁹ C.

Accepted Answer

V = kQ/r = (8.99 × 10⁹ × (−3.0 × 10⁻⁹))/0.20 (1) V = −135 V (negative because the charge is negative) (1)

Question 8

A charge of +2.0 × 10⁻⁹ C is moved from a point 0.30 m from a fixed point charge of 5.0 × 10⁻⁹ C to a point 0.10 m from it. Calculate the potential difference between the two points and the work done on the charge.

Accepted Answer

V at 0.10 m = 450 V (1) V at 0.30 m = 150 V (1) ΔV = 300 V (1) W = QΔV = 2.0 × 10⁻⁹ × 300 = 6.0 × 10⁻⁷ J (1)

Question 9

In a uniform electric field the potential changes by 200 V over a distance of 0.050 m along a field line. Calculate the electric field strength.

Accepted Answer

E = ΔV/Δd = 200/0.050 (1) E = 4000 V m⁻¹ (1)

Question 10

Two point charges, each of +4.0 μC, are placed so that each is 0.20 m from a point P. Determine the total electric potential at P.

Accepted Answer

Potential is a scalar, so the contributions add (1) V = 2 × kQ/r = 2 × (8.99 × 10⁹ × 4.0 × 10⁻⁶/0.20) (1) V = 3.60 × 10⁵ V (1)

Question 11

Determine the distance from a point charge of +12 nC at which the electric potential is 450 V. k = 8.99 × 10⁹ N m² C⁻²

Accepted Answer

V = kQ/r, so r = kQ/V (1) r = (8.99 × 10⁹ × 12 × 10⁻⁹)/450 (1) r = 0.24 m (1)

Question 12

In a vacuum tube, an electron moves from a point at a potential of +150 V to a point at a potential of +650 V. Calculate the kinetic energy the electron gains, in J and in eV. e = 1.60 × 10⁻¹⁹ C

Accepted Answer

ΔV = 650 − 150 = 500 V (1)
The electron's charge is negative, so moving to the higher potential lowers its potential energy by qΔV and that fall becomes kinetic energy: ΔE_k = eΔV = 1.60 × 10⁻¹⁹ × 500 = 8.0 × 10⁻¹⁷ J (1)
= 500 eV (1)

Question 13

A small charge is moved from point X to point Y in the field of a fixed point charge, once along a straight line and once along a much longer curved path. State and explain how the work done in the two cases compares.

Accepted Answer

The work done is the same (1); it depends only on the potential difference between X and Y, not on the path taken (1).

Question 14

A charge of +5.0 μC is placed at a point where the electric potential is 2000 V. Calculate the electric potential energy of the charge and state the work done in bringing it to this point from infinity.

Accepted Answer

E_p = QV = 5.0 × 10⁻⁶ × 2000 (1) E_p = 0.010 J (1) Work done from infinity = 0.010 J, equal to the potential energy gained (1).

Question 15

A +6.0 μC charge and a −6.0 μC charge are placed so that each is 0.30 m from a point P. Calculate the total electric potential at P and explain your result.

Accepted Answer

V = kQ₁/r + kQ₂/r (1)
V = (8.99 × 10⁹ × 6.0 × 10⁻⁶/0.30) + (8.99 × 10⁹ × (−6.0 × 10⁻⁶)/0.30) (1)
V = 0 V (1)
The two contributions are equal in magnitude but opposite in sign, and potential is a scalar, so they cancel exactly (1).

Question 16

Explain how the electric field strength at a point can be found from a graph of electric potential against distance.

Accepted Answer

E equals the magnitude of the gradient of the potential-distance graph at that point, E = ΔV/Δr (1); the field is directed from high to low potential (1). Conversely, the change in potential between two points equals the area under the field strength-distance graph (1).

Question 17

A charge of +8.0 nC is fixed at point A and a charge of −2.0 nC is fixed at point B, 0.50 m from A. Determine the distance from A of the point on the line between the charges at which the total electric potential is zero.

Accepted Answer

Potential is a scalar, so the two contributions must sum to zero: in magnitude, kQ_A/d = kQ_B/(0.50 − d) (1) 8.0(0.50 − d) = 2.0d (1) d = (8.0 × 0.50)/10.0 (1) d = 0.40 m from A (1)

Question 18

Two protons are held 1.0 × 10⁻¹² m apart in a vacuum. One is kept fixed and the other is released from rest. Calculate the speed of the released proton when it is very far from the fixed one. Assume that only the electrostatic force acts.
k = 8.99 × 10⁹ N m² C⁻², e = 1.60 × 10⁻¹⁹ C, m_p = 1.67 × 10⁻²⁷ kg

Accepted Answer

Potential at the released proton due to the fixed one: V = ke/r = 1438 V (1)
E_p = QV = 1.60 × 10⁻¹⁹ × 1438 = 2.30 × 10⁻¹⁶ J (1)
Far away V → 0, so all this potential energy becomes kinetic energy (1)
½mv² = E_p (1)
v = √(2E_p/m) = 5.2 × 10⁵ m s⁻¹ (1)

Question 19

A bird perches with both feet on a single bare overhead cable at a potential of 25 kV. Suggest why the bird is unharmed.

Accepted Answer

Both feet touch the same cable, so they are at almost exactly the same potential (1); the potential difference across the bird is therefore almost zero (1). Since W = QΔV, almost no energy is transferred through the bird and no significant current flows in it (1).