Electric Field of a Point Charge — Structured Questions (Cambridge International A-Level Physics): Exam Questions

Define electric field strength.

Fig. 1.1 shows the electric field lines between two charged spheres, A and B.

Fig. 1.1

Using Fig. 1.1, state and explain which of the spheres, A or B, has

(i) a negative charge

(ii) a greater magnitude of charge.

State the formula for the electric field strength $E$ at the surface of the sphere and state the meaning of any other symbols used.

Extract:

The formula for electric field strength at the surface of the sphere is: E = Q/(4πε₀r²) Where ε₀ is the permittivity of free space, Q is the charge on the sphere, and r is the radius of the sphere.

The field strength at the surface of sphere A is 2.6 × 10³ V cm^–1, and the field strength at the surface of sphere B is 1.3 × 10³ V cm^–1.

The magnitude of the charge stored on the surface of sphere B is 7.5 µC.

(i) Show that the radius of sphere B is about 7 cm.

(ii) Calculate the magnitude of the charge stored on the surface of sphere A. Assume both spheres have an equal radius.

State the relationship between electric field strength $E$ and the distance $r$ from a point charge.

Sphere X is a small conducting sphere that carries a charge of +2.0 × 10⁻⁹ C.

This charge may be assumed to act as a point charge situated at the centre of sphere X.

Calculate the electric field strength at a distance of

(i) 2 cm from the centre of sphere X

(ii) 10 cm from the centre of sphere X.

Extract:

Sphere X is a small conducting sphere that carries a charge of +2.0 × 10^−9^ C. This charge may be assumed to act as a point charge situated at the centre of sphere X. Calculate the electric field strength at a distance of (i) 2 cm from the centre of sphere X (ii) 10 cm from the centre of sphere X.

Coulomb constant, 1/(4πε₀) = 8.99 × 10^9 m F^–1. At 10 cm = 0.1 m: E = (8.99 × 10^9)(2.0 × 10^−9)/0.1^2 = 1800 V m^–1 (1.8 kV m^–1).

Sphere X is placed at a distance of 20 cm from sphere Y, which is identical to sphere X but carries a negative charge.

At the midpoint between the spheres X and Y:

(i) Calculate the magnitude of the electric field strength

(ii) State the direction of the electric field.

Extract:

Sphere X is a small conducting sphere that carries a charge of +2.0 × 10^−9^ C. Sphere X is placed at a distance of 20 cm from sphere Y, which is identical to sphere X but carries a negative charge. At the midpoint between the spheres X and Y: (i) Calculate the magnitude of the electric field strength (ii) State the direction of the electric field.

Sphere Y is replaced with a sphere which is positively charged and identical to sphere X.

Explain why the value of electric field strength at the midpoint between the spheres would be zero.

Fig 1.1 shows an isolated conducting sphere of radius r is placed in air. It is given a charge –Q.

Fig 1.1

State assumptions about the charge of the sphere.

The maximum field strength at the surface of the sphere before electrical breakdown (sparking) occurs is –6.2 MV m^–1. The sphere has a radius of 40 cm.

Calculate the maximum charge that can be stored on the sphere.

Extract:

Fig 1.1 shows an isolated conducting sphere of radius r is placed in air. It is given a charge –Q.

On Fig 1.2, sketch a graph to show the variation with distance d from the centre of the sphere of the electric field strength E due to the charge –Q.

Fig 1.2

Calculate the distance d from a point charge of 6.0 µC where the electric field strength is 96 kV m^–1.

Extract:

Calculate the distance d from a point charge of 6.0 µC where the electric field strength is 96 kV m^–1^.

d = 0.7496 m = 0.75 m (75 cm)

At some distance $x$, the electric field strength decreases to 10.8 kV m^–1.

Determine the value of $x$.

Extract:

Calculate the distance d from a point charge of 6.0 µC where the electric field strength is 96 kV m^–1^. (d = 0.75 m)

At some distance x, the electric field strength decreases to 10.8 kV m^–1^. Determine the value of x. (x = 2.2 m)

Fig 1.1 shows the different distances from the 6.0 µC charge in terms of d.

Fig 1.1

Identify the distance $x$ by placing an X on Fig 1.1.

An α-particle is moving directly towards a stationary gold nucleus (${}_{79}^{197}{Au}$).

The α-particle and the gold nucleus may be considered to be solid spheres with the charge and mass concentrated at the centre of each sphere.

The α-particle approaches to 0.75 × 10^–14 m from the gold nucleus where it is momentarily stationary.

Calculate the electric field strength the α-particle experiences at the point.

Extract:

An α-particle is moving directly towards a stationary gold nucleus (₇₉¹⁹⁷Au). The α-particle and the gold nucleus may be considered to be solid spheres with the charge and mass concentrated at the centre of each sphere. The α-particle approaches to 0.75 × 10⁻¹⁴ m from the gold nucleus where it is momentarily stationary. Calculate the electric field strength the α-particle experiences at the point. E = 2.02 × 10²¹ V m⁻¹ (electric field strength due to gold nucleus, from part a)

Show that the gold nucleus experiences an electric field strength that is roughly 40 times less than the α–particle.

Extract:

An α-particle is moving directly towards a stationary gold nucleus ($$_{79}^{197}A u$$). The α-particle and the gold nucleus may be considered to be solid spheres with the charge and mass concentrated at the centre of each sphere. The α-particle approaches to 0.75 × 10^–14^ m from the gold nucleus where it is momentarily stationary.

State the direction that the α–particle will travel in after it is momentarily stationary. Explain why this happens.