The potential gradient between two charged plates having separation of 0.5cm and a potential difference of 12 volts is:
Correct answer: D. 2400 NC-1
- A. 240 NC-1
- B. 24 NC-1
- C. 2.4 NC-1
- D. 2400 NC-1
Explanation
As we know, Potential gradient is given by E= -dV/dR So, 12/0.005 = 2400 V/m= 2400 N/C
Last updated
About Electric Field Intensity
Electric field intensity is the force acting per unit positive test charge, with direction set by the force on that charge. Work includes fields of point charges, vector superposition, field lines and the relation between field and electric potential. Field intensity must be distinguished from electric force and potential.
Practise Electrostatics
831 free Electrostatics MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Physics questions like this
Physics is on 13 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
A physical quantity, electric field intensity has
A proton has mass m and charge q. It is suspended in electric and gravitational field. What is the magnitude of electric field?
An electron is situated midway between two parallel plates 0.5 cm apart. One of the plates is maintained at a potential of 60 volts above the other. The force on the electron is: (e = -1.6 x 10^-19C)
Electric field at a distance of 20cm from a 4µC charge is:
Electric field intensity at a point is defined as