Questions
A point charge of is placed at Find the electric field intensity due to this charge at a point located at
detailed solution
Correct option is B
(2)r→A=3i^+4j^m, r→B=9i^+12j^m∴ r→B/A = r→B − r→A = 6i^+8j^m ⇒ r→B/A = 62+82=10m∴ EB=14πε0 Qr→B/A 2= 9× 109 × 100 × 10−6102 V/m=9000 V/mTalk to our academic expert!
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A charged oil drop is suspended in a uniform field of 3x10 4 v/m so that it neither falls nor rises .The charge on the drop will be (take mass of the charge =9.9 x10-15 kg and g = 10m /s2
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