AP Physics B and C- Electrostatics: Additional Multiple Choice Practice Questions on Field and Potential

As promised in the post dated 20^th June 2008, we will discuss a few more multiple choice questions on electric field and potential.

The following questions are meant for AP Physics B as well as AP Physics C:

(1) Three positive charges each of magnitude q are arranged symmetrically along the circumference of a circle of radius r. The electric field at the centre of the circle has magnitude

(a) 3q /4πε₀r²

(b) √3q /4πε₀r²

(c) q/4πε₀r²

(d) (1+√2)q/4πε₀r²

(e) zero

The electric field at any point is the force acting on unit positive charge. The three charges arranged symmetrically along the circle will exert (on unit positive charge placed at the centre) forces of the same magnitude with their lines of action inclined at 120º. The resultant force is therefore zero. Hence the field at the centre of the circle is zero.

(2) A thin ring of radius R is uniformly charged with a total amount of positive charge Q. The electric potential at a point P on the axis of the ring at distance x from the centre of the ring is

(a) Q /4πε₀√( r²+x²)

(b) Qr /2ε₀√( r²+x²)

(c) Q /4πε₀ ( r²+x²)

(d) Qr /4πε₀√( r²+x²)

(e) Qx/4πε₀√( r²+x²)

The distance of point P from the charges on the ring is √( r²+x²) so that the electric potential at P is Q /4πε₀√( r²+x²) given in option (a).

(3) Two identical conducting spheres, each of radius R, carry charges +Q and –Q/2 respectively. When the distance between their centres is d (d>2R), the attractive force between them is F. The spheres are now brought into contact and then separated to the initial distance d. The force between the spheres is now

(a) attractive and of magnitude F/4

(b) repulsive and of magnitude F/4

(c) attractive and of magnitude F/8

(d) repulsive and of magnitude F/8

(e) zero

The initial attractive force between the charged spheres is F = (1/4πε₀) (Q²/2d²)

When the spheres are brought into contact they share the charges. The total charge on the system of the two spheres is the sum of the initial charges and is equal to +Q/2. Since the spheres are identical, they share the charges equally so that the charge on each sphere is +Q/4.

The final force between the spheres is therefore repulsive and of magnitude (1/4πε₀) (Q²/16d²) = F/8.

The following practice questions are for AP Physics C aspirants only:

(1) Two identical thin non-conducting spherical shells of radius R carry equal negative charges of magnitude Q sprayed uniformly on their surfaces. The distance AB between their centres is 4R. The electric field at a point P (Fig.) at distance R/2 from the centre A of the first sphere is

(a) Q/64πε₀R² directed towards right

(b) Q/64πε₀R² directed towards left

(c) Q/49πε₀R² directed towards left

(d) Q/49πε₀R² directed towards right

(e) zero

The electric field at P is the resultant of the fields due to the two shells. But since the point P is inside the first shell (with centre at A), the field due to the first shell is zero. Therefore the field at P is solely due to the second shell. The distance of the centre of the second shell from P is 4R – (R/2) = 7R/2. Therefore the magnitude of the field at P is given by

E = (1/4πε₀)[Q/(7R/2)²] = Q/49πε₀R²

Since the charge on the sphere is negative, the field at P is directed towards right [Option (d)].

(2) If the spherical shells in the above question are made of a conductor, the electric field at P will be

(a) Q/64πε₀R² directed towards right

(b) Q/64πε₀R² directed towards left

(c) Q/49πε₀R² directed towards left

(d) Q/49πε₀R² directed towards right

(e) zero

Since the point P is inside the first shell, the field at P due to the charges on the first shell is zero as in the above question. The field at P due to the charges on the second shell also is zero since the conducting shell provides electrostatic screening of the electric field produced by the second shell. Therefore the correct option is (e).

(3) A thin walled cubical box of side 10 cm has a constant potential of 4 V on its surface. There are no charges inside the box. The potential at the centre of the box is

(a) 2V

(b) 4V

(c) zero

(d) 8V

(e) 2V/3

Since there are no charges inside the box, there cannot be any potential gradient on that account. Since the potential everywhere on the surface of the box is 4 V, there is no potential gradient inside the box on this account also. To put this in a different manner, there is no electric field inside the box and hence the potential at the centre is 4 V itself.

(4) A charge Q is divided into two parts as q and (Q – q) so that for any given separation between q and (Q – q), the electrostatic interaction between q and (Q – q) is maximum. The ratio Q/q is

(a) 8

(b) 6

(c) 4

(d) 3

(e) 2

The electrostatic force between the two parts is repulsive and is given by

F = q(Q – q) /4πε₀r²

This will be maximum when dF/dq = 0

Therefore, –2q + Q = 0 from which q = Q/2.

The ratio Q/q = 2

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You know that a soap bubble has an excess of pressure inside because of surface tension. What will happen to the excess of pressure on charging the bubble? Because of the repulsion between the charges, the bubble will expand, thereby reducing the excess of pressure.