logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

The sides of a rectangle ABCD are 15 cm and 5 cm as shown in Figure. Point charges of $– 5 \mu C$ and $+ 2 \mu C$ are placed at the corners D and B respectively. Calculate electric potential at A and C. Also, calculate work done in carrying a charge of $3 \mu C$ from A to C.

Can you answer this question?
 
 

1 Answer

0 votes
As is clear from Figure.
$V_A= \large\frac{1}{4 \pi \in_0} \bigg[ \large\frac{2 \times 10^{-6}} {15 \times 10^{-2}} \frac{5 \times 10^{-6}}{5 \times 10^{-2}}\bigg]= \frac{9 \times 10^9 \times 10^{-6}}{10^{-2}} \bigg[\large\frac{2}{15} -1\bigg]$$=7.8 \times 10^5\;V$
$V_C= \large\frac{1}{4 \pi \in_0} \bigg[ \large\frac{2 \times 10^{-6}} {5 \times 10^{-2}} \frac{5 \times 10^{-6}}{15 \times 10^{-2}}\bigg]= \frac{9 \times 10^9 \times 10^{-6}}{10^{-2}} \bigg[\large\frac{2}{5} -\frac{1}{3}\bigg]$$=0.6 \times 10^5\;V$
Work done in carrying a charge $q = 3 \times 10^{–6} C$ from A to C.
$w= q(V_C-V_A)=3 \times 10^{-6} [0.6 \times 10^5 -(-7.8 \times 10^{5} )]=2.52 \;J$
answered Jun 8, 2014 by meena.p
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...