Two large horizontal metal plates are separated by 4 mm. The lower plate is at a potential of –80 V.

Question 13

Two large horizontal metal plates are separated by 4 mm. The lower plate is at a potential of –80 V.

Which potential should be applied to the upper plate to create an electric field of strength

60 000 V m^-1 upwards in the space between the plates?

A –320 V                     B –160 V                     C +160 V                     D +320 V

Reference: Past Exam Paper – November 2016 Paper 11 & 13 Q32

Solution:

Answer: A.

Electric field strength E = V / d

Where V is the potential difference between the plates

E = V / d

V = E × d = 60 000 × 4×10^-3

Potential difference V = 240 V

The potential difference between the plates is 240 V. The bottom plate is at a potential of –80 V. The difference between the lower plate and the upper plate should be 240 V.

The upper plate be       EITHER greater than –80 V by 240 V

OR it could be smaller than –80 V by –240 V

But since the electric field is upwards, the top plate must be at a lower potential than the bottom plate (since electric field is drawn from high(er) potential to low(er) potential).

Therefore a potential of (–80 – 240 =) –320 V is required on the top plate [A is correct].

We need to consider the direction of the field. Merely calculating correct potential difference would make C a possible answer. But when the direction of the field is considered, only A can be correct.

It may be helpful to remember that the field gives the direction of the force on a positive charge. For a positive charge to experience a force upwards, the top plate must be at a more negative potential than the bottom plate.