Two point charges 5 micro coulomb and -5 micro coulomb are located at A and B seperated by 0.2 m in vaccum . What is the electric feild at the midpoint of the line joining the two charges
Answers
Answer:
0.45 J
Explanation:
Explanation:
Given in the question that there are two charges A and B
Value of first charge = +5 uC
Value of second charge = +6 uC
Distance between them = 12 cm
Distance after the charge B have move 2 cm toward charge A = 12-2 = 10cm
Work done in moving the charge is equal to the gain in the electrostatic potential of the system U
U1 = \frac{1}{4\pi e }\frac{q1q2}{r1}
4πe
1
r1
q1q2
U2 = \frac{1}{4\pi e }\frac{q1q2}{r2}
4πe
1
r2
q1q2
WD = U2 - U1
WD=\frac{q1q2}{4\pi e}(\frac{1}{r2}-\frac{1}{r1})WD=
4πe
q1q2
(
r2
1
−
r1
1
)
Plug in the values
\frac{(5*10^-6)(6*10^-6)}{4\pi e}(\frac{1}{10*10^-2}-\frac{1}{12*10^-2})
4πe
(5∗10
−
6)(6∗10
−
6)
(
10∗10
−
2
1
−
12∗10
−
2
1
)
= 0.45 J
Answer:
idint know about this answer
Explanation:
hiiiiiiiii hiiiiï