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# Using Eqs. 2.27 and 2.30, find the potential at a distance

ISBN: 9780321856562 45

## Solution for problem 25P Chapter 2

Introduction to Electrodynamics | 4th Edition

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Introduction to Electrodynamics | 4th Edition

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Problem 25P

Using Eqs. 2.27 and 2.30, find the potential at a distance z above the center of the charge distributions in Fig. 2.34. In each case, compute E = −∇V, and compare your answers with Ex. 2.1, Ex. 2.2, and Prob. 2.6, respectively. Suppose that we changed the right-hand charge in Fig. 2.34a to −q; what then is the potential at P? What field does that suggest? Compare your answer to Prob. 2.2, and explain carefully any discrepancy.

Eqs. 2.27

Eqs. 2.30

Reference example 2.1

Find the electric field a distance z above the midpoint between two equal charges (q), a distance d apart (Fig. 2.4a).

Reference example 2.1

Find the electric field a distance z above the midpoint of a straight line segment of length 2L that carries a uniform line charge λ (Fig. 2.6).

Reference prob.2.6

Find the electric field a distance z above the center of a flat circular disk of radius R (Fig. 2.10) that carries a uniform surface charge σ.What does your formula give in the limit R→∞? Also check the case z ≫R.

Reference figure 2.10

Reference prob.2.2

Find the electric field (magnitude and direction) a distance z above the midpoint between equal and opposite charges (±q), a distance d apart (same as Example 2.1, except that the charge at x = +d/2 is −q).

Reference example 2.1

Find the electric field a distance z above the midpoint between two equal charges (q), a distance d apart (Fig. 2.4a).

Step-by-Step Solution:

Step 1 of 5</p>

Part a

We are required to calculate the potential at the point  as given in figure 2.3 (a).

Let us have a look at the following figure.

Therefore, the potential at the point  at a distance z above is,

Step 2 of 5</p>

Part b

From equation 2.30,

Let us check the following figure.

From the equation,

The potential due to uniform line charge,

This is the expression for potential due to the line charge.

Step 3 of 5</p>

Part c

From equation 2.30,

Here,  is surface charge density.

Therefore, the total charge

From the above equation,

This is the required expression for potential due to uniform surface charge.

Calculations of

Step 4 of 5

Step 5 of 5

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