The electric field intensity between two uniformly, oppositely charged parallel plates is:
Correct answer: C. σ/ε0
- A. ε0 / σ
- B. 2σ/ε0
- C. σ/ε0
- D. σ/E
Explanation
The electric field intensity between two uniformly, oppositely charged parallel plates can be given by: σ/ε0 where σ is the surface charge density on the plates and ε0 is the permittivity of free space. The surface charge density σ is defined as the total charge Q divided by the area A of one of the plates: σ = Q/A The electric field intensity E0 between the plates is related to the surface charge density by the equation: E = σ/ε As it is a numerical only one option D is correct.
Report an error
The more specific you are, the faster it gets fixed. A source beats an opinion.
Prefer email? [email protected]
About Electrostatics
Electrostatics describes forces between charges through Coulomb's law, electric fields and field intensity, then uses electric potential to measure energy per unit charge. Capacitors store charge and electrical energy, with capacitance depending on geometry and dielectric material, not simply on the amount of charge present.
Practise Electrostatics
831 free Electrostatics MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Discussion
Stuck on an option, or know a faster way to get there? Ask or explain it here.
No comments yet. Be the first to explain this one.
Exams that ask Physics questions like this
Physics is on 16 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
More Electrostatics questions
In equations E=.-V/r Negative sign indicates that:
If a slab of dielectric constant Er=2 is inserted between, the parallel plates of a charged 20 uF capacitor. Its capacitance will:
The electric field at a point situated at a distance d from a straight charged conductor is:
If 2uF and 4 uF capacitors connected in series. Their equivalent capacitance will be:
Unit, joule per coulumb is equal to:
A proton of charge 1.6 x 10^-19 C is moved through a potential difference of 20 volt The energy gained by the proton is: