^{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) 3*q* /4πε_{0}*r*^{2}

(b) √3*q* /4πε_{0}*r*^{2}

(c) *q*/4πε_{0}*r*^{2}

(d) (1+√2)*q*/4πε_{0}*r*^{2}

(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* /4πε_{0}√(_{ }*r*^{2}+*x*^{2})

(b) *Qr* /2ε_{0}√(_{ }*r*^{2}+*x*^{2})

(c) *Q* /4πε_{0} (_{ }*r*^{2}+*x*^{2})

(d) *Qr* /4πε_{0}√(_{ }*r*^{2}+*x*^{2})

(e)** ***Qx/*4πε_{0}√(_{ }*r*^{2}+*x*^{2}) ** **

The distance of point P from the charges on the ring is √(_{ }*r*^{2}+*x*^{2}) so that the electric potential at P is *Q* /4πε_{0}√(_{ }*r*^{2}+*x*^{2}) 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>*2*R*), 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/*4πε_{0}) (*Q ^{2}/*2

*d*

^{2})

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/*4πε_{0}) (*Q ^{2}/*16

*d*

^{2}) =

*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 4

*R*. The electric field at a point P (Fig.) at distance

*R/*2 from the centre A of the first sphere is

(a) *Q*/64πε_{0}*R*^{2} directed towards right

(b) *Q*/64πε_{0}*R*^{2} directed towards left

(c) *Q*/49πε_{0}*R*^{2} directed towards left

(d) *Q*/49πε_{0}*R*^{2} 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 4*R* – (*R/2*) = 7*R*/2. Therefore the magnitude of the field at P is given by

*E =* (1/4πε_{0}*)*[*Q/*(7*R/*2)^{2}] = *Q*/49**πε _{0}R^{2}**

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*/64πε_{0}*R*^{2} directed towards right

(b) *Q*/64πε_{0}*R*^{2} directed towards left

(c) *Q*/49πε_{0}*R*^{2} directed towards left

(d) *Q*/49πε_{0}*R*^{2} 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*) /4πε_{0}*r*^{2}

This will be maximum when d*F*/d*q* = 0

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

The ratio *Q/q* = 2

** ** ** ** ** ** ** ** ** ** ** **

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