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two parallel plates of area 5.68*10^-4 m^2 have equal and opposite charges of 8.38*10^-11 C placed on them. what is the electric field between the plates

Sagot :

Answer: unit n/c

Explanation: This should be the answer because I just asked my Physics Teachers I hope this is correct

Hope This Helps :)

The electric field between the plates will be 0.1667×10⁵ C.

What is an electric field?

An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.

The electric field formula is as follows:

E = F /Q

Where,

The electric field is denoted by the letter E.

F is Electric force

The charge is Q.

Given data;

Q (Charge)= 8.38 × 10¯¹¹ C ,

A (Area between the plates) = 5.68 × 10¯⁴ m²,

ε₀(permeability constant) = 8.85 × 10¯¹² C²/Nm²

[tex]\rm E= \frac{Q}{K \epsilon_0 } \\\\ \rm E= \frac{8.38 \times 10^{-11}}{5.68 \times 10^{-4} \times 8.85 \times 10^{-12} } \\\\ E= 0.1667 \times 10^5 \ N C^{-1}[/tex]

Hence,the electric field between the plates will be 0.1667×10⁵ C.

To learn more about the electric field refer to the link;

https://brainly.com/question/26690770

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