Question

A thin, horizontal, 18-cm-diameter copper plate is charged to -3.8 nC. Assume that the electrons are uniformly distributed on the surface. Part A) What is the strength of the electric field 0.1 mm above the center of the top surface of the plate? Express your answer to two significant figures and include the appropriate units. Part B) What is the strength of the electric field 0.1 mm below the center of the bottom surface of the plate? Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer:

Part a)

$E = 8436.7 N/C$

Part b)

$E = 8436.7 N/C$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = \frac{Q}{A}$

$Q = -3.8 nC$

$A = \pi(0.09)^2$

$A = 0.025 m^2$

$\sigma = \frac{-3.8\times 10^{-9}}{0.025}$

$\sigma = -1.5 \times 10^{-7} C/m^2$

now the electric field is given as

$E = \frac{-1.5 \times 10^{-7}}{2(8.85 \times 10^{-12})}$

$E = 8436.7 N/C$

Part b)

Now since the electric field is required at same distance on other side

so the field will remain same on other side of the plate

$E = 8436.7 N/C$

Answer 2

The electric field due to the given large sheet is 8436.4 N/C. It is an area in space where a charged particle experiences a force.

What is an electric field?

It is an area in space where a charged particle experiences a force. The  field due to the large sheet can be given as,

$E = \dfrac \rho {2\epsilon_0}$

Since,

$\rho = \dfrac QA$

So,

$E = \dfrac Q {2A \epsilon_0}$

Where,

$Q$ - charge = - 3.8 nC

$A$ - area  = $\pi r^2 = 3.14 \times 0.09^2 = 0.025 \ \rm \ m^2$

$\epsilon_0$ - electric permeability = $8.85\times 10^{-12}$

Put the values in the formula,

$E = \dfrac{- 3.8} {2\times 0.025 \times8.85\times 10^{-12} }\\\\E = 8436.4 \rm\ N/C$

Therefore, the electric field due to the given large sheet is 8436.4 N/C.

Learn more about the electric field:

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