Question

50 points question ...,,if you know than please solve 13 question
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Answer 1

at a distances of 41.4 (or)41.7cm from the first charge this is the answer

Answer 2

Explanation:

q1=5×10−19C

q2=20×10−19C

Electric field at P is 0

∴E1=E2

(2−x)2Kq1=x2Kq2

q1x2=q2(2−x)2

3×10−19×x2=420×10−19(2−x)2

x2=4(2−x)2

x2=[2(2−x)]2

x=2(2−x)

x=4−2x

2x=