If we did all things we are capable of, we would literally astound ourselves.

– Thomas A. Edison

Saturday, June 21, 2008

Electrostatics: MCQ on Charged Spherical Drops

Often questions involving the calculation of the potential of a drop obtained by the combination of a number of identical charged spherical droplets are seen in college entrance question papers. If n identical small drops each of radius r carrying charge q coalesce to form a single large drop of radius R, we have

R = n1/3 r

[You will get this by equating the volumes: n×(4/3)πr3 = (4/3)πR3]

The potential V on the surface of each small drop is given by

V = (1/4πε0)(q/r)

The total charge on the larger drop is nq. Therefore, the potential V’ on its surface is

V’ = (1/4πε0)(nq/R) = (1/4πε0)(nq/n1/3r) since R = n1/3 r

Therefore, V’ = n2/3V

The electric field E on the surface of each small drop is given by

E = (1/4πε0)(q/r2)

The electric field E’ on the surface of the large drop is given by

E’ = (1/4πε0)(nq/R2) = (1/4πε0)(nq/n2/3r2 ) since R = n1/3 r

Therefore, E’ = n1/3E

If you can remember the above expressions for the electric potential and field on the larger drop, multiple choice questions involving them can be answered in no time. But it is always more rewarding to remember the basic things so that you can calculate the required quantities in all situations.

The calculation of V’ in the above form itself appeared as a multiple choice question in Kerala Engineering Entrance 2008 question paper.

Now answer the following multiple choice question involving charged drops. You may try them yourself and then check with the given solution. Here are the questions:

(1) Sixty four identical drops each charged by q coulomb to potential V volt coalesce to form a single large drop. The charge and potential on the large drop are respectively

(a) 64q, 64V

(b) 64q, V

(c) 8q, 16V

(d) 16q, 16V

(e) 64q, 16V

The charge on the large drop is the sum of the charges on the small drops and is equal to 64q.

As shown above, the potential on the large drop is V’ = n2/3V = 642/3V =16V [Option (e)].

(2) Two identical soap bubbles A and B are uniformly charged with the same amount of charge. But the charge on A is positive where as the charge on B is negative. (The electrical interaction between the bubbles is negligible). Because of charging, the excess of pressure inside

(a) both bubbles will increase

(b) both bubbles will decrease

(c) both bubbles will remain unchanged

(d) A will increase and that inside B will decrease

(e) B will increase and that inside A will decrease

Because of the repulsive force between like charges, the bubbles will expand and hence the excess of pressure inside both bubbles will decrease.

You will find many useful posts on different branches of Physics at your level at apphysicsresources.blogspot.com and at physicsplus.blogspot.com. The essential equations to be remembered in the section, ‘Electric field and Potential’ can be found here.

Thursday, June 12, 2008

Kerala Medical Entrance (KEAM) 2008 Questions from Newton’s Laws of Motion

The following questions which appeared in Kerala Medical Entrance 2008 question paper are of the type popular among question setters:

(1) Two blocks of masses 7 kg and 5 kg are placed in contact with each other on a smooth surface. If a force of 6 N is applied on the heavier mass, the force on the lighter mass is

(a) 3.5 N

(b) 2.5 N

(c) 7 N

(d) 5 N

(e) 6 N

The common acceleration a of the system of masses is given by

a = Driving force/Total mass moved = 6 N/(7+5) kg = 0.5 ms–2

The force on the lighter mass = Ma = 5×0.5 = 2.5 N.

(2) A body of mass 60 kg suspended by means of three strings P, Q and R as shown in the figure is in equilibrium. The tension in the string P is

(a)130.9g N

(b) 60g N

(c) 50g N

(d) 103.9g N

(e) 109g N

Let us denote the tensions in the three strings by P, Q and R. The concurrent forces P, Q and R keep the common meeting point of the strings in equilibrium so that they can be represented by the three sides of a triangle taken in order as shown in the adjoining figure. From this right angled triangle,

Q/P = tan 30º where Q = weight of the mass M = Mg =60g newton.

[This equation is often written as W/H = tan θ where W is the weight and H is the horizontal force and is known as tangent law].

Substituting for Q, we have 60g/P = 1/√3, from which P = 60g×√3 = 103.9g N.