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Two charges 'q1' and 'q2' of magnitude 1...

Two charges `'q_1'` and `'q_2'` of magnitude `10^(-8) C` and `- 10^(-8) C`, respectively, are placed 0.1 m apart. Calculate the electric field at points A and B as shown in Fig.

Text Solution

Verified by Experts

To calculate the electric field at point A:

Here `q_1 = +10^(8)C`
`q_2 = -10^(8)C`
`E_1 and E_2` are electric fields at point A due to `q_2 and q_2`
`E_1 =1/(4piin_0)(q_1)/((0.05)^2)`
and `E_2 =1/(4piin_0)(q_2)/((0.05)^2)`
`:. E_1 and E_2` are in same direction.
`:.` Net electric field at `A=E+E_1+E_2`
`E=1/(4piin_0)(q_1)/((0.05)^2)+1/(4piin_0)(q_2)/((0.05)^2)`
`1/(4piin_0)[(q_1)/((0.05)^2)+(q_2)/((0.05)^2)]`
`= 9 xx10^9[(10^(-8))/((0.05)^2)+(10^(-8))/((0.05)^2)]`
`9 xx10^(9)xx(2xx10^(-8))/((0.05)^2)`
`= (18xx10^(11))/(0.05 xx0.05)=180/25xx10^4`
`=7.2 xx10^(4) NC^(-1)` along BA
To calculate the electric field at point B

`E_1` = Electric due to charge `q_1` at point B
`E_1 =1/(4piin_0)(q_1)/((0.05)^2)`
`E_2` = Electric due to charge `q_2` at point B
`E_2 =1/(4piin_0)(q_2)/((0.15)^2)`
`:. E_1 and E_2` are in opposite direction
`:.` Net electric field at B `B=E=(E_1-E_2)` along AB.
`E=1/(4piin_0)q_2/((0.05)^2)-1/(4piin_0)q_2/((0.15)^2)`
`E=1/(4piin_0)[q_1/((0.05)^2)-q_2/((0.15)^2)]`
`= 9 xx10^9 [(10^(-8))/((0.05)^2)-10^(-8)/((0.15)^2)]`
`= 9 xx10^9 [(10^(-4))/25-(10^(-4))/255]`
`= 9 xx10^9 xx10^(-4)[1/25-1/25]`
`= 9 xx10^5xx[(9-1)/225]`
`= 9 xx10^5 xx8/225`
`=72/200xx10^5=720/220xx10^4`
`E=3.27 xx10^(4) NC^(-1)` along AB
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