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# Two point charges q_A = 3 μC and q_B = –3 μC are located 20 cm apart in vacuum. What is the electric field at the midpoint O of the line AB joining the two charges?

Q 1.8(a): Two point charges $q_{A}=3 \mu C$ and $q_{B}=-3 \mu C$ are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges?

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Given, AB = 20 cm

Since, O is the midpoint of the line AB.

AO = OB = 10 cm = 0.1m

The electric field at a point caused by charge q, is given as,

$E = \frac{kq}{r^2}$

Where, q is the charge, r is the distance between the charges and the point O

k = 9x109 N m2 C-2

Now,

Due to charge at A, Electric field at O will be $E_{A}$ and in the direction AO.

$E_{A} =\frac{ 9\times10^9 \times 3\times10^{-6}}{0.1^2}$

Similarly the Electric field at O due to charge at B,  also in the direction AO

$E_{B} =\frac{ 9\times10^9 \times (-3\times10^{-6})}{0.1^2}$

Since, both the forces are acting in the same direction, we can add their magnitudes to get the net electric field at O:

E' = $E_{A}$ + $E_{B}$ = 2E (Since their magnitudes are same)

$E' =2\times \frac{ 9\times10^9 \times 3\times10^{-6}}{0.1^2} = 5.4\times 10^4 N/C$ along the direction AO.

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