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Two point charges q_{A} = 3 μC and q_{B} = −3 μC 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?

**(b)** If a negative test charge of magnitude 1.5 × 10^{−9} C is placed at this point, what is the force experienced by the test charge?

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#### Solution

**(a)** The situation is represented in the given figure. O is the mid-point of line AB.

Distance between the two charges, AB = 20 cm

∴ AO = OB = 10 cm

Net electric field at point O = E

Electric field at point O caused by +3μC charge,

`"E"_1 = (3 xx 10^-6)/(4piin_0("AO")^2) = (3 xx 10^-6)/(4piin_0(10 xx 10^-2)^2) "N"//"C" `along OB

Where,

`in_0` = Permittivity of free space

`1/(4piin_0) = 9 xx 10^9 "Nm"^2"C"^-2`

Magnitude of electric field at point O caused by −3μC charge,

`"E"_1 = (-3 xx 10^-6)/(4piin_0("OB")^2) = (3 xx 10^-6)/(4piin_0(10 xx 10^-2)^2) "N"//"C" `along OB

`therefore "E" = "E"_1 + "E"_2`

= `2xx[(9xx10^9)xx(3xx10^-6)/((10xx10^-2)^2)]` ......[since the value of `"E"_1` and `"E"_2` are same, the value is multiplied with 2]

= 5.4 × 10^{6} N/C along OB

Therefore, the electric field at mid-point O is 5.4 × 10^{6} N C^{−1} along OB.

**(b) **A test charge of amount 1.5 × 10^{−9} C is placed at mid-point O.

q = 1.5 × 10^{−9} C

Force experienced by the test charge = F

∴ F = qE

= 1.5 × 10^{−9} × 5.4 × 10^{6}

= 8.1 × 10^{−3} N

The force is directed along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.

Therefore, the force experienced by the test charge is 8.1 × 10^{−3} N along with OA.

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