Question

A particle with charge +6.70 nC is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the A particle with a charge of 7.50 nC is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right. After it has moved a distance of 6.50 cm , the additional force has done an amount of work equal to 7.60×10−5 J and the particle has kinetic energy equal to 3.00×10−5 J .
(a) What work was done by the electric force?

(b) What is the potential of the starting point with respect to the end point?

(c) What is the magnitude of the electric field?

Answer 1

Correct question:

A particle with a charge of 7.50 nC is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right. After it has moved a distance of 6.50 cm , the additional force has done an amount of work equal to 7.60×10−5 J and the particle has kinetic energy equal to 3.00×10−5 J .

(a) What work was done by the electric force?

(b) What is the potential of the starting point with respect to the end point?

(c) What is the magnitude of the electric field?

Answer:

(a) work done by the electric force is −4.6 × 10⁻⁵ J

(b) change in potential is −6.133 x 10³ J

(c) the magnitude of the electric field is 9.435 x 10⁴ V/m

Explanation:

Given;

charge of the particle, q = +7.50 nC

distance moved by the particle, d = 6.50 cm

work done by the additional force, W = 7.60 × 10⁻⁵ J

kinetic energy of the particle, K.E = 3.00 × 10⁻⁵ J.

Part (A)

work done by the electric force

$W = W_e_f + W_f ------equation(i)\\\\W = K.E_f + K.E_i -------equation(ii)$

where;

$W_e_f$ is work done by the electric force

$W_f$ is  work done by additional force

$K.E_f$ is final kinetic energy

$K.E_i$ is initial kinetic energy, this zero since the particle was released from rest

$W_e_f + W_f = K.E_f\\\\W_e_f = K.E_f - W_f$

       = 3.00 × 10⁻⁵ J - 7.60 × 10⁻⁵ J

       = − 4.6 × 10⁻⁵ J

Part (B)

change in potential from the starting point with respect to the end point, is calculated as follow,

$W_e_f$ = QΔV

ΔV = $W_e_f$ / Q

ΔV  = (− 4.6 × 10⁻⁵) / (7.5 x 10⁻⁹)

ΔV = − 6.133 x 10³ J

Part (C)

Electric field is given as;

E = ΔV / d

E = (− 6.133 x 10³) / (6.5 x 10⁻²)

E = − 9.435 x 10⁴ V/m

thus, the magnitude of the electric field is 9.435 x 10⁴ V/m

Answer 2

Answer:

a) 4.6*10^{-5} J

b) 6133.33 J/C

c) 94358.9 C

Explanation:

(a) We have that the change in the kinetic energy equals the net work over the charge. Hence we have

$\Delta E_K=W-W_E$

where Ek is the kinetic energy, W is the work of the external force and WE is the work done by the electric field. By replacing we obtain:

$W_E=W-\Delta E_k=7.60*10^{-5}J-3*10^{-5}J=4.6*10^{-5}J$

(b) The potential difference is computed by using:

$\Delta V=\frac{W_E}{q}=\frac{4.6*10^{-5}J}{7.5*10^{-9}C}=6133.33\frac{J}{C}$

(c) With the work done by the electric force we can calculate the Electric field. By using the following formula we obtain:

$W_E=qEd\\\\E=\frac{W_E}{qd}=\frac{4.6*10^{-5}J}{(7.5*10^{-9}C)(6.50*10^{-2}m)}=94358.9\frac{N}{C}$

where we have used d=6.5cm=6.5*10^-9m and q=7.5nC=7.5*10^-9C

hope this helps!!