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)?

a)

6.33 x 10⁸ N/C

Direction : Towards negative charge.

b)

1.11125 x 10²⁰ m/s²

Direction : Towards positive charge.

Explanation:

a)

= magnitude of negative charge = 25 x 10⁻⁶ C

= magnitude of positive charge = 50 x 10⁻⁶ C

= distance of negative charge from point P = 0.02 m

= distance of positive charge from point P = 0.08 m

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

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

Net electric field at P is given as

Direction:

Towards the negative charge.

b)

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

= 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

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

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

EA = 5.625 x 10^8 N/C

Electric field at P due to the charge at B

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 = 1.11 x 10^20 m/s^2

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