Question

the electronic charge is 1.6*10^-19.if a body carries a negative charge of 9.6*10-19c .the number of excess electrons of the body is

Answer 1

Explanation:

Given The electronic charge is 1.6*10^-19.if a body carries a negative charge of 9.6*10-19c .the number of excess electrons of the body is

• Given electronic charge is e = 1.6 x 10^-19 and a body carries a negative charge q = 9.6 x 10^-19 C
• So we need to find the number of excess electrons that is n
• So n = q / e
•       = 9.6 x 10^- 10 / 1.6 x 10^- 19
• Or n = 6 x 10^9
• Therefore the number of excess electrons in the body will be 6 x 10^9

Reference link will be

https://brainly.in/question/3122398

Answer 2

The number of excess electrons of the body is 6

Given:

Electronic charge = e = 1.6 × 10⁻¹⁹ C

Negative charge = Q = 9.6 × 10⁻¹⁹ C

Explanation:

The quantisation of charge is given by the formula:

Q = n × e

Where,

n = Number of excess electrons

On substituting the values, we get,

9.6 × 10⁻¹⁹ = n × (1.6 × 10⁻¹⁹)

n = (9.6 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹)

n = 9.6/1.6

∴ n = 6