Physics, asked by pramod56, 1 year ago

Urgent help plzzzzzzzzzzzzzzz....
Solve this

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Answers

Answered by kgayatri683

Answer:

Let charge at B be Q

EA at B due to A = k (9×10^-10)/9 = 9×10^9 × 10^-10

= 0.9N/C

EC at B due to C = k (16×10^-10)/16 = 9×10^9×10^-10

= 0.9N/C

EA = EC

Enet = root EA^2 + EC^2 + 2 EA EC cos90

= root 2EA^2 + 0

= root 2 × EA

= 1.414 × 0.9 = 1.2726 N/C

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pramod56: but I can't understand madam

kgayatri683: Where u r not getting

kgayatri683: Will u first confirm me the answer

pramod56: 1.29N/C

kgayatri683: ok let me check error

pramod56: OK

shrutikushwaha96: You can't add the electric field intensity simply because their direction is different

pramod56: I too think of it...

kgayatri683: hmm I know I used vector law to add them

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