Q#21 (Past Exam Paper – June 2019 Paper 42 Q6)
(a) State what is meant by electric potential at a point. [2]
(b) Two parallel metal plates A and B are held a distance d apart in a vacuum, as illustrated in Fig. 6.1.
Fig. 6.1
Plate A is earthed and plate B is at a potential of +V0.
Point P is situated in the centre region between the plates at a distance x from plate B.
The potential at point P is V.
On Fig. 6.2, show the variation with x of the potential V for values of x from x = 0 to
x = d.
Fig. 6.2
[3]
(c) Two isolated solid metal spheres M and N, each of radius R, are situated in a vacuum. Their centres are a distance D apart, as illustrated in Fig. 6.3.
Fig. 6.3
Each sphere has charge +Q.
Point P lies on the line joining the centres of the two spheres, and is a distance y from the centre of sphere M.
On Fig. 6.4, show the variation with distance y of the electric potential at point P, for values of y from y = 0 to y = D.
Fig. 6.4
[4]
[Total: 9]
Solution:
(a) The electric potential at a point is the work done in moving unit positive charge from infinity to the point.
(b)
{The electric field strength is the negative of the potential gradient.
That is, in a potential-distance graph, the gradient gives the negative of the field strength.
Electric field strength = - potential gradient
The electric field strength between parallel plates is constant. So, the graph should be a straight line.
Electric field strength has a positive value. So, the gradient should be negative so that the field strength is positive.
Therefore, the graph is a straight line with negative gradient.
From the diagram,
When x = 0, potential = +V0 (0, +V0)
When x = d, potential = 0 (d, 0)}
(c)
{The electric potential inside a conductor is constant. That is, from y = 0 to y = R and from y = D-R to y = D, the potential is constant – it is a horizontal line.
Potential V = Q / 4πϵ0r
Since both spheres have the same charge +Q, the magnitudes of the potentials at the surfaces of the spheres is the same.
The potential decreases as we move away from one sphere and increases again as we approach the other sphere. The potential is minimum at the mid-point of the curve as they have the same charge.
Since the charges are both positive, the potential is also positive.}
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