Questions
Equipotential surfaces are shown in figure. Then the electric field strength will be
detailed solution
Correct option is C
Using dV=-E→·dr→ ⇒ΔV=-E.Δrcosθ ⇒E=-ΔVΔrcosθ ⇒E=-(20-10)10×10-2cos120° =-1010×10-2-sin30°=-102-1/2=200 V/mDirection of E be perpendicular to the equipotential surface i.e., at 120° with x-axis.Talk to our academic expert!
Similar Questions
Two parallel plates separated by a distance of 5 mm are kept at a potential difference of 50 V. A particle of mass 10-15 kg and charge 10-11 C enters in it with a velocity 107 m/s. The acceleration of the particle will be
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