Question

An electric field of 7.50×105 V/m is desired between two parallel plates, each of area 45.0 cm2 and separated by 2.45 mm of air. What charge must be on each plate?

Answer 1

To determine the charge on each plate for a desired electric field in a capacitor with parallel plates, use the formula Q = E x A x ε_0. Convert the area to m^2, insert all values including the permittivity of free space, and solve for Q.

Step-by-step explanation:

To find the charge required on each of the parallel plates to create a specific electric field, you can use the relationship between the electric field (E), the charge (Q) on the plates, and the permittivity of free space (ε0). The electric field between two parallel plates is given by the equation:

E = Q / (ε0 × A)

where E is the electric field, Q is the charge on each plate, A is the area of the plates, and ε0 is the permittivity of free space (ε0 = 8.854 x 10−12 C2/N·m2). Rearranging this for Q gives:

Q = E × A × ε0

First, convert the area from cm2 to m2 by multiplying by (10−4)2. Then, plug in the values:

Q = (7.50 × 105 V/m) × (0.45 × 10−4 m2) × (8.854 × 10−12 C2/N·m2)

Calculate Q to find the charge required on each plate.

Answer 2

Charge,
`q=2.98* 10^(-8)\ C`

Step-by-step explanation:

It is given that,

Value of electric field,
`E=7.5* 10^5\ V/m`

Area of parallel plates,
`A=45\ cm^2=0.0045\ m^2`

Distance between two parallel plates, d = 2.45 mm = 0.00245 m

For a parallel plate capacitor, the capacitance is given by :

`C=(\epsilon_oA)/(d)`.......(1)

Since,
`E=(V)/(d)`

V = E . d ............(2)

And
`C=(q)/(V)`.....(3)

From equation (1), (2) and (3) we get :

`(q)/(V)=(\epsilon_oA)/(d)`

`q=\epsilon_o EA`

`q=8.85* 10^(-12)\ F/m* 7.5* 10^5\ V/m* 0.0045\ m^2`

`q=2.98* 10^(-8)\ C`

So, the charge on the each plate is
`2.98* 10^(-8)\ C`. Hence, this is the required solution.