Question

A solid conducting sphere of radius 5.00 cmcarries a net charge. To find the value of the charge, you measure the potential difference VAB=VA−VB between point A, which is 8.00 cm from the center of the sphere, and point B, which is a distance r from the center of the sphere. You repeat these measurements for several values of r >8cm. When you plot your data as VAB versus 1/r, the values lie close to a straight line with slope -18.0 V⋅m.

Required:
What does your data give for the net charge on the sphere?

Answer 1

Answer:

The value is $q = 2 *10^{-9} \ C$

Explanation:

From the question we are told that

The radius is $r = 5.00 \ cm = 0.05 \ m$

The distance of point A from the center is $a = 8.0 \ cm = 0.08 \ m$

The distance of point B from the center is $b = r$

The slope is $s = -18.0 \ V\cdot m.$

Generally the difference in potential between A and B is mathematically represented as

$V_{AB} = V_A -V_B = k * q [\frac{1}{a} - \frac{1}{b} ]$

Let consider the position where $b = r = \infty$

So

$V_{AB} = V_A -V_B = k * q [\frac{1}{a} - \frac{1}{ \infty} ]$

Here k is the coulombs constant with value $k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

=> $V_{AB} = V_A -V_B = 9*10^{9} * q [\frac{1}{0.08} - 0 ]$

=> $V_{AB} = V_A -V_B = 9*10^{9} * q [\frac{1}{0.08} - 0 ]$

=> $V_{AB} = 1.125 *10^{11} q \ V$

Generally the slope is mathematically represented as

$s = \frac{V_{AB}}{\frac{1}{r} }$

At the position of A i.e r = a

$s = \frac{V_{AB}}{\frac{1}{a} }$

=> $-18.0 = \frac{ 1.25 *10^{11} q}{\frac{1}{0.08} }$

=> $-18.0 * 12.5 = 1.125 *10^{11} q$

=> $q = 2 *10^{-9} \ C$

Answer 2

The net charge is the sum of negative and positive charges. The net charge of the given sphere is $2\times 10^{-9} \rm \ C$.

Net charge:

It can be calculated by the formula,

$V_{AB} = A_a-V_b\\\\V_{AB} = k \times q[\frac 1a - \frac 1b]$

Where,

$k$ - Coulomb's constant  

$a$ - distance from the center = 0.08

$b$ - distance of B from center = r

Assume, r = infinite

Then,

$V_{AB} = 9\times10^9\times q[\frac 1{0.08} - \frac 1{0}]\\\\V_{AB} = 1.125\times 10^{11}\times q \rm \ V$

At the position of A, the slope,

$s = \dfrac {V_{AB}}{\frac 1{a}}$

Put the values,

$-18= \dfrac {1.125\times 10^{11}\times q}{\frac 1{0.08}}\\\\q = 2\times 10^{-9}$

Therefore, the net charge of the given sphere is $2\times 10^{-9} \rm \ C$.

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