Question

1. What is the electric field strength at a point 50 cm from q = +5.00nC?

a. 180 n/c
b. 360 n/c
c. 540 n/c
d. 720 n/c

2. A 1.65 NC charge with a mass of 1.5 × 10-15 kg experiences an acceleration of 6.33 × 107 m/s2 in an electric field. What is the magnitude of the electric field?

a. 14.96 N/C
b. 29.35 N/C
c. 57.55 N/C
d. 2.67 × 10-19 N/C​

Answer 1

Answer:

1. the test point on the -x axis, 50 cm from the charge q=-5.00

Answer 2

Answer:

1) Option a) 180 N/C

2) Option c) 57.55 N/C

Explanation:

1) Given,

Distance, d = 50 cm

• Convert 50 cm into m
• 1 cm = 0.01 m
• 50 cm = 0.5 m

Charge, Q= 5 nC

• Convert 5 nC into C
• 1 nC = 10⁻⁹ C
• 5 nC = 5 × 10⁻⁹ C

We can find the electric field, E, created by a point charge by using this equation:

$E = k\frac{Q}{d^{2} }$

Here, k is electrostatic constant, 9 × 10⁹ Nm²/C.

$E = 9 * 10^{9} \frac{5 * 10^{-9} }{(0.5)^{2} }$

$E = \frac{45 }{0.25}$

E = 180 N/C

2) Given,

Charge, Q = 1.65 nC = 1.65 × 10⁻⁹ C

Mass, m = 1.5 × 10⁻¹⁵ Kg

Acceleration, a = 6.33 × 10⁷ m/s²

Magnitude of the electric field is given by:

$E = \frac{ma}{Q}$

$E = \frac{1.5 * 10^{-15}* 6.33 * 10^{7} }{1.65 * 10^{-9} }$

E = 5.755 × 10 N/C

E = 57.55 N/C

RESULT: 1) Option a) 180 N/C

               2) Option c) 57.55 N/C