Two point charges qA = 3 μC and qB = –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?
Answers
(a) Given:
qA = 3 μC = 3 × 10‒6 C
qB = ‒3 μC = ‒ 3 × 10‒6 C
r = 10 cm (point O is in middle)
We know, electric field is given by formula
E = Q/(4πϵo r2 )
Electric field due to charge A (by using above formula)
|EA| = 2.7 × 106 N / C
Electric field due to charge B
|EB| = 2.7 × 106 N/C opposite to the direction of EA.
Since both the fields have the same direction, the electric field at O will be summation of above field.
So, |ENet| = 5.4 × 106 N/C towards B.
(b) Given:
qo = 1.5 × 10–9 C
|ENet| = 5.4 × 106 N/C towards B
By using the formula
F = q E
We have,
F = (qo) (|ENet|)
= (1.5 × 10–9 C) × (5.4 × 106 N/C)
= 8.1 × 10‒3 towards charge qB (because direction of electric field is towards B).
Answer:
b ) A testcharge of amount 1.5 × 10−9 C is placed atmid-point O.
q= 1.5 × 10−9 C
Forceexperienced by the test charge = F
∴F= qE
= 1.5 × 10−9 × 5.4 × 106
= 8.1 × 10−3 N
Theforce is directed along line OA. This is because the negative testcharge is repelled by the charge placed at point B but attractedtowards point A.
Therefore,the force experienced by the test charge is 8.1 × 10−3N along OA.