“*Your worth consists in what you are and not in what you have*”

–Thomas A. Edison

Multiple choice as well as free response practice questions on electrostatics (with solution) were posted on this blog on many occasions. You can access them either by clicking on the label ‘electrostatics’ below this post or by trying a search for ‘electrostatics’, making use of the search box on this page. In both cases you would need to click on the ‘older posts’ and ‘newer posts’ buttons to access *all *posts. Today we will discuss a few more multiple choice practice questions in this section.

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

(1) In a region of the atmosphere near the earth’s surface there exists an electric field of intensity 100 Vm^{–1}directed vertically downwards. Consider a line segment of length 4 m inclined at 60º with the vertical (fig.). What is the potential difference between the points A and B?

(a) 200 V

(b) 100 V

(c) 60 V

(d) 50 V

(e) 10 V

Since the electric field is 100 Vm^{–1}and is directed *vertically*, a difference in height of 1 m corresponds to a potential difference of 100 V. The difference in height between the points A and B is AB cos60º = 4×½ = 2 m.

Therefore, the potential difference between points A and B is 200 V.

(2) A positive point charge Q located at point A produces an electric field of 50 NC^{–1} at point B. If an equal and opposite charge (– Q) is placed *additionally* at the point B, the electric field and electric potential at the mid point of the line AB are respectively

(a) 400 NC^{–1}, 50 V

(b) 400 NC^{–1}, 0 V

(c) 200 NC^{–1}, 50 V

(d) 200 NC^{–1}, 0 V

(e) 100 NC^{–1}, 0 V

The electric fields due to the positive charge at A and the negative charge at B are directed from A to B and they will get added at the mid point of the line AB. The magnitudes of these fields are equal and is 200 NC^{–1} each, since the field is *quadrupled* when the distance is *halved* (in accordance with the inverse square law). The resultant electric field at the mid point of the line AB is 400 NC^{–1}.

The electric potential at the mid point of the line AB is *zero* since the positive potential produced by the positive charge is canceled by the equal negative potential produced by the negative charge [Option (b)].

(3) Two conducting spheres A and B have radii *R* and 3*R* respectively. They are given charges *Q* and 3*Q* respectively. If they are connected by a copper wire,

(a) a charge *Q/*2 will flow from B to A

(b) a charge *Q/*2* *will flow from B to A

(c) a charge *Q *will flow from B to A

(d) a charge *Q *will flow from A to B

(e) no charge will flow through the wire

The potentials of A and B are the same. Hence no charge will flow through the wire.

[Note that the potentials of A and B are (1/4πε_{0})(*Q/R*) and (1/4πε_{0})(3*Q/*3*R*), which are equal].

(4) Point charges 1μC, 2μC, –3μC and 4μC are arranged in free space at the corners A, B, C and D of a square ABCD of side √2 m. If 1/4πε_{0} is approximately equal to 9×10^{9} Nm^{2}C^{–2}, the electric potential at the centre of the square is

(a) 1.8×10^{4} V

(b) 3.6×10^{4} V

(c) 2.7×10^{4} V

(d) –2.7×10^{4} V

(e) –6.3×10^{4} V

The diagonal of the square has length 2 m and the distance of the centre from the corners is 1 m. The positive charges of total value 7μC will produce (at the centre) a total positive potential of (1/4πε_{0}) (7×10^{–6 }/1) volt = (9×10^{9})^{ }(7×10^{–6}) volt = 6.3×10^{4} volt.

[Note that the potential due to a point charge *Q* at distance *r* is (1/4πε_{0})(*Q/r*)].

The negative charge of value –3μC will produce (at the centre) a negative potential of (1/4πε_{0}) (–3×10^{–6 }/1) volt = (9×10^{9})^{ }(–3×10^{–6}) volt = –2.7×10^{4} volt.

Therefore, the electric potential at the centre of the square is (6.3×10^{4 }–2.7×10^{4}) volt = 3.6×10^{4} volt.

(5) The potential on an equipotential surface is *V* volt. The work done in moving a charge *q *through a distance *d* along the equipotential surface is

(a) directly proportional to *V*, *q* and *d*

(b) directly proportional to *V* and *q* but inversely proportional to *d*

(c) directly proportional to *V* and *d* but inversely proportional to *q*

(d) directly proportional to *q* and *d* but inversely proportional to *V*

(e) independent of *V*, *q* and *d*

The correct option is (e) since the work done in moving a charge along the equipotential surface is zero.

[Since the surface is equipotential, you don’t require a force to move the charge along the surface].

The following questions are meant for ** AP Physics C** aspirants; but AP Physics B aspirants also can ‘enjoy’ them:

(6) Two identical thin rings A and B of radius *R* are arranged coaxially with a separation *R* between their centres. Ring A carries *positive* charge 2*Q* where as ring B carries *positive *charge *Q*. What is the external work requied to move a *positive* charge *q* from the centre of A to the centre of B?

(a) (1/4πε_{0})[*Q/*(√2 *R*)]

(b) (1/4πε_{0})(2*Q/**R*)

(c) (1/4πε_{0})[*Q/*(√2 *R*) –* Q/R*]

(d) (1/4πε_{0})[*Q/R*_{ }– *Q/*(√2 *R*)]

(e) (1/4πε_{0})[*Q/*(√2 *R*) +* Q/R*]

With reference to the adjoining figure, the electric potential *V*_{1}* *at the centre of ring A is given by

*V*_{1} = (1/4πε_{0})[2*Q/R *+ *Q/*(√2 *R*)]

[The first term is due to the charge on A while the second term is due to the charge on B].

The electric potential *V*_{2}* *at the centre of ring B is given by

*V*_{2} = (1/4πε_{0})[*Q/R* + 2*Q/*(√2 *R*)]

[The first term is due to the charge on B while the second term is due to the charge on A].

The potential difference *V *between the centres of A and B is given by

*V* = *V*_{1 }the centre of A* V*_{2} = (1/4πε_{0})[2*Q/R*_{ }–* Q/R *+ *Q/*(√2 *R*) – 2*Q/*(√2 *R*)]

Or, *V* = (1/4πε_{0})[*Q/R*_{ }– *Q/*(√2 *R*)]

Since the centre of A is at a higher positive potential, a positive charge *q *placed at the centre of A will move by itself towards the centre of B. So the work to be done by the external agency is *negative* and the answer to the question is – *V* and we have

– *V = *(1/4πε_{0})[*Q/*(√2 *R*) –* Q/R*], as given in option (c).

(7) A total charge (positive) of 1.11×10^{–9 }C is sprayed non uniformly on a non conducting ring of radius 1 m. If **E** represents the electric field at any point in space due to this charge distribution and **r** = 0 represents the centre of the ring, what is the value (in joule per coulomb) of the line integral ∫– **E.dr** from r = ∞ to r = 0? ^{}

(a) zero

(b) ∞

(c) 1

(d) 10

(e) 100

The quantity given by the line integral from infinity to the centre of the ring is the work done in bringing unit positive charge from infinity to the centre of the ring. This is the electric potential at the centre of the ring and is equal to (1/4πε_{0})(*Q/R*) where *Q* is the total charge on the ring and *R* is the radius of the ring.

Since 1/4πε_{0} = 9×10^{9} Nm^{2}C^{–2} (very nearly), we obtain the answer as 9×10^{9}×1.11×10^{–9}/1 = 10, very nearly.

Well, you will find a very useful post in this section here.

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