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QThe electric field intensity and electric potential at a certain distance from a point charge are 32 N/C and 16 J/C, respectively. What is the distance from the charge?

ID: #11149 Electrostatics 113 views
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#11149Q ID
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ElectrostaticsTopic
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  • A 50 m ✔ ✖
  • B 0.5 m ✔ ✖
  • C 10 m ✔ ✖
  • D 7 m ✔ ✖
Correct Answer: Option B

Explanation

The electric field intensity E at a distance r from a point charge q is given by E = k*q/r^2, where k is the electrostatic constant. The electric potential V at that distance is given by V = k*q/r. Given that E = 32 N/C and V = 16 J/C, we can equate the two expressions and solve for r. Therefore, 32 N/C = (16 J/C) / r. Rearranging the equation, we have r = (16 J/C) / 32 N/C = 0.5 m.

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