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(a) Twelve equal charges, q,are situated at the comers of a regular 12-sided polygon (for instance, one on each numeral of a clock face). What is the net force on a test charge Qat the center?

(b) Suppose oneof the 12 q'sis removed (the one at "6 o'clock"). What is the force on Q?Explain your reasoning carefully.

(c) Now 13 equal charges, q,are placed at the comers of a regular 13-sided polygon. What is the force on a test charge Qat the center?

(d) If one of the 13 q'sis removed, what is the force on Q?Explain your reasoning.

Short Answer

Expert verified

(a)Total force on the test charge is zero.

(b)Due to the charge at 12 o clock, force is given by F=140qQr2

(c)The force on the test charge at the centre of the polygon is zero.

(d)If any one charge from corner of the polygon is removed, then the net force due to the remaining 12 charge is given by following equation.F=140qQr2

Step by step solution

01

Describe the given information

The polygon having 12 sides, have the charges present at each of its corner. A test charge is present at the center of the polygon having 12 sides. The force and electric filed at the test charge is to be evaluated.

02

Define the Gauss divergence theorem and stokes theorem

Electric force exerted by chargeqon charge Qis proportional to theproduct of the two charge and inversely proportional to the square of thedistance between them as,

F=140qQr2.

03

Prove expression in part (a)

If two charge qand Qare placed at a distance rapart. Then electric force exerted by charge qon charge Qis proportional to the product of the two charge and inversely proportional to the square of the distance between them as,

F=140qQr2 鈥︹︹.. (1)

If the two charges have same sign that is positive or negative then the force between them is repulsive and directed radially outward.

As the charges are lying at the corners of polygon and charges are alike, and also placed at the same distance from the test charge at the center. So, force between each corner charge and the test charge is same and repulsive in nature.

All the charge apply the same force on the test charge in opposite direction. So, total force on the test charge is zero.

04

Prove expression in part (b)

The force on the test charge at the center, for the condition, the charge which is at 6 o clock is removed is obtained as follows.

There is only charge present exactly opposite to the test charge, which is due to the charge at 12 o clock. Now the force at the test charge for the given condition is given by

F=140qQr2

05

Prove expression in part (c)

If the polygon has 13 sides, then also the force on the charge at the center is 0. This is because of the symmetry of the system. When all the charges are at the same distance from the test charge, the force on the center charge becomes zero.

It can be understood via a circular ring with uniformly distributed charge on it. The force on the center of the ring is zero due to symmetry. This is because the ring is made up of combination of even or odd number of infinite parts.

Thus, the force on the test charge at the center of the polygon is zero.

06

Prove expression in part (d)

If there are 13 charges on the corners of a polygon, then electric field due to the 13 charges at the corners at the test charge will act in different directions, due to different position of the test charge with respect to corner charges.

If net force is computed on the test charge, it will sum up to zero.

As an example, electric force due to 13 charge, at the center is 0. This is because, the force on the test charge due to the other 12 corner charges is balanced by 13th charge.

So, if any one charge from corner of the polygon is removed, the resultant force will not be equal to 0. Then, in this condition, the net force due to the remaining 12 charge is given by following equation.

F=140qQr2

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Most popular questions from this chapter

Suppose the plates of a parallel-plate capacitor move closer together by an infinitesimal distance, as a result of their mutual attraction.

(a) Use Eq. 2.52 to express the work done by electrostatic forces, in terms of the fieldE, and the area of the plates, A.

(b) Use Eq. 2.46 to express the energy lost by the field in this process.

(This problem is supposed to be easy, but it contains the embryo of an alternative derivation of Eq. 2.52, using conservation of energy.)

Check that Eq. 2.29 satisfies Poisson's equation, by applying the Laplacian and using Eq. 1.102.

Consider two concentric spherical shells, of radiiaand b.Suppose the inner one carries a charge q ,and the outer one a charge -q(both of them uniformly distributed over the surface). Calculate the energy of this configuration, (a) using Eq. 2.45, and (b) using Eq. 2.47 and the results of Ex. 2.9.

Imagine that new and extraordinarily precise measurements have revealed an error in Coulomb's law. The actual force of interaction between two point charges is found to be

F=14蟺蔚0q1q2r2(1+r)e(r)r^

where 位 is a new constant of nature (it has dimensions of length, obviously, and is a huge number鈥攕ay half the radius of the known universe鈥攕o that the correction is small, which is why no one ever noticed the discrepancy before). You are charged with the task of reformulating electrostatics to accommodate the new discovery. Assume the principle of superposition still holds.

a. What is the electric field of a charge distribution 蟻 (replacing Eq. 2.8)?

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c. Find the potential of a point charge q鈥攖he analog to Eq. 2.26. (If your answer to (b) was "no," better go back and change it!) Use 鈭 as your reference point.

d. For a point charge q at the origin, show that

SE.da+12V痴诲蟿=10q

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e. Show that this result generalizes:

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