Q. 9 A sphere of radius R has charge ... [FREE SOLUTION]

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Chapter 23: Q. 9 (page 652)

A sphere of radius R has charge Q. The electric field strength at distance $r > R$ is $E_{i}$ . What is the ratio $E_{f} / E_{i}$ of the final to initial electric field strengths if
(a) Q is halved,
(b) R is halved, and
(c) r is halved (but is still $> R$ )? Each part changes only one quantity; the other quantities have their initial values

Short Answer

Expert verified

(a) The ratio of the final to initial electric field strengths, if Q is halved, is $1 / 2$ .

(b) The ratio of the final to initial electric field strengths if radius R is halved is $1$ .

Step by step solution

Given information (part a)

$C h a r g e = Q_{i} = Q C h a r g e Q_{f} = \frac{Q}{2} d_{i} = r + R 鈮� r (S i n c e r > R) d_{f} = r + R 鈮� r (S i n c e r > R)$

Explanation (part a)

$s p h e r e of radius E_{i} = \frac{Q_{i}}{4 蟺蔚_{0} d_{i}^{2}} S p h e r e o f r a d i u s E_{f} = \frac{Q_{f}}{4 蟺 e_{0} d_{f}^{2}} 鈬� \frac{E_{f}}{E_{i}} = \frac{\frac{Q_{f}}{4 蟺 e_{0} r^{2}}}{\frac{Q}{4 蟺 e_{0} r^{2}}} 鈬� \frac{E_{f}}{E_{i}} = \frac{Q}{2} \frac{1}{Q} 鈬� \frac{E_{f}}{E_{i}} = \frac{1}{2}$

Given information (part b)

$S p h e r e o f r a d i u s r_{i} = R C h a r g e Q_{i} = Q C h a r g e Q_{f} = Q d_{i} = (R + r) d_{f} = \frac{R}{2} + \frac{R}{2} + r = (R + r)$

Explanation (part b)

$s p h e r e of radius E_{i} = \frac{Q_{i}}{4 蟺蔚_{0} d_{i}^{2}} S p h e r e o f r a d i u s E_{f} = \frac{Q_{f}}{4 蟺 e_{0} d_{f}^{2}} 鈬� \frac{E_{f}}{E_{i}} = \frac{\frac{Q_{f}}{4 蟺蔚_{0} (R + r)^{2}}}{\frac{Q_{i}}{4 蟺蟿_{0} (R + r)^{2}}} 鈬� \frac{E_{f}}{E_{i}} = \frac{Q}{1} \frac{1}{Q} 鈬� \frac{E_{f}}{E_{i}} = \frac{1}{1} 鈬� \frac{E_{f}}{E_{i}} = 1$

Given information (part c)

$Sphereofradius r_{i} = R C h a r g e Q_{i} = Q S p h e r e o f r a d i u s r_{f} = R C h a r g e Q_{f} = Q d_{i} = (R + r) 鈮� r d_{f} = \frac{r}{2} + R 鈮� \frac{r}{2}$

Explanation (part c)

$s p h e r e of radius E_{i} = \frac{Q_{i}}{4 蟺蔚_{0} d_{i}^{2}} S p h e r e o f r a d i u s E_{f} = \frac{Q_{f}}{4 蟺 e_{0} d_{f}^{2}} 鈬� \frac{E_{f}}{E_{i}} = \frac{\frac{Q_{f}}{4 蟺系_{0} {(\frac{r}{2})}^{2}}}{\frac{Q_{i}}{4 蟺系_{0} (r)^{2}}} 鈬� \frac{E_{f}}{E_{i}} = \frac{Q}{1} \frac{4}{Q} 鈬� \frac{E_{f}}{E_{i}} = 4$