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

The strength of the electric field 0.1mm above the center of the top surface of the plate is -0.0067 N/C. The strength of the electric field 0.1mm below the center of the bottom surface of the plate is also -0.0067 N/C.

Step-by-step explanation:

To find the strength of the electric field above the center of the top surface of the plate, we can use the formula for electric field due to a uniformly charged plate:

E = σ/(2ε0)

Where E is the electric field, σ is the charge density (charge per unit area), and ε0 is the permittivity of free space.

Part A)

Since the charge is evenly distributed on the surface, the charge density can be calculated by dividing the total charge by the area:

σ = Q/A

where Q is the total charge and A is the surface area of the plate. Now we can substitute this value into the formula to find the electric field:

E = Q/(2ε0A)

Plugging in the given values, the electric field is:

E = (-3.8nC)/(2ε0 × (π × (9cm)^2))

E = -0.0067 N/C

Therefore, the strength of the electric field 0.1 mm above the center of the top surface of the plate is -0.0067 N/C.

Part B)

Similarly, to find the strength of the electric field below the center of the bottom surface of the plate, we can use the same formula:

E = σ/(2ε0)

The charge density is the same, but now we are interested in the electric field below the plate. Since the electric field is a vector quantity, the direction will change when we move to the bottom surface. Therefore, the strength of the electric field 0.1 mm below the center of the bottom surface of the plate is also -0.0067 N/C.

Answer 2

Part a)

`E = 8436.7 N/C`

Part b)

`E = 8436.7 N/C`

Step-by-step explanation:

Part a)

Electric field due to large sheet is given as

`E = (\sigma)/(2\epsilon_0)`

`\sigma = (Q)/(A)`

`Q = -3.8 nC`

`A = \pi(0.09)^2`

`A = 0.025 m^2`

`\sigma = (-3.8* 10^(-9))/(0.025)`

`\sigma = -1.5 * 10^(-7) C/m^2`

now the electric field is given as

`E = (-1.5 * 10^(-7))/(2(8.85 * 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`