Question

Two point charges are separated by a distance of 10.0 cm. One has a charge of -25 μC and the other +50 μC. (a) Determine the direction and magnitude of the electric field at a point P between the two charges that is 2.0 cm from the negative charge. (b) If an electron (mass = 9.11 x 10-31 kg) is placed at rest at P and then released, what will be its initial acceleration (direction and magnitude)?

Answer 1

Step-by-step explanation:

qA = - 25 x 10^-6 C

qB = 50 x 10^-6 C

AP = 2 cm

BP = 8 cm

(a)

Electric field at P due to the charge at A

`E_(A)=(kq_(A))/(AP^(2))`

`E_(A)=(9* 10^(9)* 25* 10^(-6))/(0.02^(2))`

EA = 5.625 x 10^8 N/C

Electric field at P due to the charge at B

`E_(B)=(kq_(B))/(BP^(2))`

`E_(A)=(9* 10^(9)* 50* 10^(-6))/(0.08^(2))`

EB = 0.70 x 10^8 N/C

The resultant electric field at P due to both the charges is

E = EA+ EB = (5.625 + 0.7) x 10^8

E = 6.325 x 10^8 N/C towards left

(b) mass of electron, m = 9.1 x 10^-31 kg

Let a be the acceleration of electron

Force on electron, F = charge of electron x electric field

F = q x E

`a = (qE)/(m)`

`a = (1.6* 10^(-19)* 6.325* 10^(8))/(9.1* 10^(-31))`

a = 1.11 x 10^20 m/s^2

Answer 2

a)

6.33 x 10⁸ N/C

Direction : Towards negative charge.

b)

1.11125 x 10²⁰ m/s²

Direction : Towards positive charge.

Step-by-step explanation:

a)

`Q_(1)` = magnitude of negative charge = 25 x 10⁻⁶ C

`Q_(2)` = magnitude of positive charge = 50 x 10⁻⁶ C

`r_(1)` = distance of negative charge from point P = 0.02 m

`r_(2)` = distance of positive charge from point P = 0.08 m

Magnitude of electric field at P due to negative charge is given as

`E_(1) = (kQ_(1))/(r_(1)^(2) ) = ((9*10^(9))(25*10^(-6)))/(0.02^(2) ) = 5.625*10^(8) N/C`

Magnitude of electric field at P due to positive charge is given as

`E_(2) = (kQ_(2))/(r_(2)^(2) ) = ((9*10^(9))(50*10^(-6)))/(0.08^(2) ) = 0.703125*10^(8) N/C`

Net electric field at P is given as

`E = E_(1) + E_(2)\\E = 5.625*10^(8) + 0.703125*10^(8) \\E = 6.33*10^(8) N/C`

Direction:

Towards the negative charge.

b)

`m` = mass of the electron placed at P = 9.31 x 10⁻³¹ C

`Q_(1)` = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

Acceleration of the electron due to the electric field at P is given as

`a = (qE)/(m)\\ a = ((1.6*10^(-19))(6.33*10^(8)))/((9.11*10^(-31)))\\a = 1.11125*10^(20) ms^(-2)`

Direction: Towards the positive charge Since a negative charge experience electric force in opposite direction of the electric field.