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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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