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Question

Two point charges Q1=400μC and Q2=100μC are kept fixed, 60 cm apart in vacuum. Find intensity of the electric field at midpoint of the line joining Q1 and Q2.

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Solution

Step 1: Electric field at midpoint O due to both charges
As, Distance between two charges, d=60cm and O is the mid point.
So, AO=BO=d2=30cm

At point O, electric field due to point charge kept at A,
E1=14πϵ0×Q1r2=9×109×400×106(30×102)2[in the direction of AO]

At point O, electric field due to point charge kept at B,
E2=14πϵ0Q2r2=9×109×100×106(30×102)2[in the direction of BO]
Step 2: Resultant Electric field at midpoint O
So, resultant electric field E=E1+E2

E=9×109(30×102)2×106×(400100)

=9×109×106×300900×104

=3×107N/C [in the direction of AO]

2106258_595864_ans_aa9d2eeaf7da424892362e29b433081f.png

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