Question

A thin conducing plate 2.3 m on a side is given a total charge of −20.0 µC. (Assume the upward direction is positive.) (a) What is the electric field (in N/C) 1.0 cm above the plate? (Indicate the direction with the sign of your answer.)

Answer 1

Answer:

The electric field is $-2.14\times10^{5}\ N/C$

Explanation:

Given that,

Distance = 2.3 m

Charge $q= -20.0\ \muC$

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{\sigma}{2\epsilon_{0}}$

Where, $\sigma=\dfrac{Q}{A}$

Q = charge

A = area

Put the value into the formula

$E=\dfrac{Q}{2A\epsilon_{0}}$

$E=\dfrac{-20.0\times10^{-6}}{2\times2.3\times2.3\times8.85\times10^{-12}}$

$E=-213599.91\ N/C$

$E=-2.14\times10^{5}\ N/C$

Negative sign shows the direction of the electric field.

The direction of electric field is toward the plates.

Hence, The electric field is $-2.14\times10^{5}\ N/C$