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Chapter 17: Q30P (page 473)

(III) Two equal but opposite charges are separated by a distance d, as shown in Fig. 17鈥�41. Determine a formula for \({{\bf{V}}_{{\bf{BA}}}}{\bf{ = }}{{\bf{V}}_{\bf{B}}}{\bf{ - }}{{\bf{V}}_{\bf{A}}}\)for points B and A on the line between the charges situated as shown.

FIGURE 17-41 Problem 30

Short Answer

Expert verified

The required formula for electric potentialis\({V_{{\rm{BA}}}} = 2kq\left[ {\frac{{2b - d}}{{\left( {d - b} \right)b}}} \right]\).

Step by step solution

01

Understanding the electric potential

The electric potential energy per unit charge at any point is termed as the electric potential. It relies on the charge and the distance of the point from the charge.

The expression for electric potential is given as:

\(V = \frac{{kQ}}{r}\)

Here, k is the Coulomb鈥檚 constant, Q is the charge and r is the distance.

02

Evaluation of the electric potential at point A and B

The relation of electric potential at Ais given by,

\({V_{\rm{A}}} = \frac{{kq}}{b} + \frac{{k\left( { - q} \right)}}{{d - b}}\)

Here, kis the electric constant and bis the distance from the end point.

The relation of electric potential at Bis given by,

\({V_{\rm{B}}} = \frac{{kq}}{{d - b}} + \frac{{k\left( { - q} \right)}}{b}\)

03

Evaluation of the electric potential difference between charges

The relation of electric potential difference is given by,

\({V_{{\rm{BA}}}} = {V_{\rm{B}}} - {V_{\rm{A}}}\)

On plugging the values in the above relation.

\(\begin{aligned}{V_{{\rm{BA}}}} &= \left( {\left( {\frac{{kq}}{{d - b}} + \frac{{k\left( { - q} \right)}}{b}} \right) - \left( {\frac{{kq}}{b} + \frac{{k\left( { - q} \right)}}{{d - b}}} \right)} \right)\\{V_{{\rm{BA}}}} &= kq\left( {\frac{1}{{d - b}} - \frac{1}{b} - \frac{1}{b} + \frac{1}{{d - b}}} \right)\\{V_{{\rm{BA}}}} &= 2kq\left( {\frac{1}{{d - b}} - \frac{1}{b}} \right)\\{V_{{\rm{BA}}}} &= 2kq\left( {\frac{{2b - d}}{{\left( {d - b} \right)b}}} \right)\end{aligned}\)

Thus, the potential difference between points A and B is \({V_{{\rm{BA}}}} = 2kq\left( {\frac{{2b - d}}{{\left( {d - b} \right)b}}} \right)\).

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Most popular questions from this chapter

Question: How does the energy stored in an isolated capacitor change if (a) the potential difference is doubled, or (b) the separation of the plates is doubled?

In the dynamic random access memory (DRAM) of a computer, each memory cell contains a capacitor for charge storage. Each of these cells represents a single binary bit value of 鈥�1鈥� when its 35-fF capacitor \(\left( {{\bf{1}}\;{\bf{fF = 1}}{{\bf{0}}^{{\bf{ - 15}}}}\;{\bf{F}}} \right)\) is charged at 1.5 V, or 鈥�0鈥� when uncharged at 0 V.

(a) When fully charged, how many excess electrons are on a cell capacitor鈥檚 negative plate?

(b) After charge has been placed on a cell capacitor鈥檚 plate, it slowly 鈥渓eaks鈥� off at a rate of about \({\bf{0}}{\bf{.30}}\;{\bf{fC/s}}\). How long does it take for the potential difference across this capacitor to decrease by 2.0% from its fully charged value? (Because of this leakage effect, the charge on a DRAM capacitor is 鈥渞efreshed鈥� many times per second.) Note: A DRAM cell is shown in Fig. 21鈥�29.

Question: Three charges are at the corners of an equilateral triangle (side l) as shown in Fig. 17鈥�45. Determine the potential at the midpoint of each of the sides. Let \[{\bf{V = 0}}\] at \[{\bf{r = }}\infty \].

FIGURE 17鈥�45 Problem 75.

(II) The work done by an external force to move a \( - {\bf{6}}{\bf{.50}}\;{\bf{\mu C}}\) charge from point A to point B is \({\bf{15}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}\;{\bf{J}}\). If the charge was started from rest and had \({\bf{4}}{\bf{.82 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}\;{\bf{J}}\)of kinetic energy when it reached point B, what must be the potential difference between A and B?

Question: (I) Write the binary number 1010101010101010 as a decimal number.
See all solutions

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