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Q.

A charge of 10–10C is placed at the origin. The electric field at (1, 1) m due to it (in NC–1) is

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a

20j¯

b

i¯+j¯

c

0.452(i¯+j¯)

d

4.52(i¯+j¯)

answer is B.

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Detailed Solution

q=1010c,r¯=i+j|r|=2;E¯=14πε0qr3×r¯=9×109×101022(i+j)=0.452(i+j)

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