CBSE Class 12 · 2003 · ElectrostaticsVery Short Answer

Two point charges

q_{A} = + 3 μ C

and

q_{B} = - 3 μ C

are located 20 cm apart in vacuum.
(i) Find the electric field at the mid-point of the line AB joining the two charges.
(ii) If a negative test charge of magnitude

1.5 \times 1 0^{- 9} C

is placed at the centre, find the force experienced by the test charge.

Solution

The mid-point is at

r = 0.10 m

from each charge.
Both charges give a field pointing from the

+

charge towards the

-

charge, so the two contributions add:

E = 2 \cdot \frac{1}{4 π ε _{0}} \frac{q}{r ^{2}}

E = 2 \times 9 \times 1 0^{9} \times \frac{3 \times 1 0 ^{- 6}}{( 0.10 ) ^{2}}

E = 5.4 \times 1 0^{6} N C^{- 1} (from A towards B)

Force on the test charge:

F = ∣ q_{0} ∣ E = 1.5 \times 1 0^{- 9} \times 5.4 \times 1 0^{6}

F = 8.1 \times 1 0^{- 3} N

Being negative, the test charge is pushed opposite to

E

, i.e. from B towards A.

CBSE Class 12 Board · 2003 · Subjective

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CBSE Class 12 · 2003 · ElectrostaticsVery Short Answer

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