Question

If mass of electione = 9.1 x 10-31 kg.
charges on the eletrone = 1.6x 10 19 coalomb.
and g = 9.8 m/Second ². then the intensity
of the electric field recpuited to balanced
the weight of an electrone is -​

Answer 1

Answer:

• The Electric Field intensity (E) is 55.73 × 10⁽⁻¹²⁾ N/C.

Given:

1. Mass of Electron (M) = 9.1 × 10⁻³¹ Kg
2. Charge of Electron (q) = 1.6 × 10⁻¹⁹ C
3. Acceleration due to gravity (g) = 9.8 m/s²

Explanation:

$\rule{300}{1.5}$

Electric field intensity:

It is the measure of the electrostatic force per unit charge at a point is called Electric Field Intensity.

Expression:

• E = F / q

S.I Units:

• N / C (or) Newton/coulomb

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From the formula we know,

⇒ E = F / q

Where,

• E Denotes Electric field intensity
• F Denotes Applied Force.
• q Denotes Charge.

Now,

⇒ E = F / q

Substituting the values,

⇒ E = M g / q

∵ [ F = M g = W ]

⇒ E = ( 9.1 × 10⁻³¹ × 9.8 ) / 1.6 × 10⁻¹⁹

⇒ E = 89.18 × 10⁻³¹ / 1.6 × 10⁻¹⁹

⇒ E = 89.18 × 10⁻³¹ × 10¹⁹ / 1.6

⇒ E = 89.18 × 10⁽⁻³¹⁺¹⁹⁾ / 1.6

⇒ E = 89.18 × 10⁽⁻¹²⁾ / 1.6

⇒ E = 55.73 × 10⁽⁻¹²⁾

⇒ E = 55.73 × 10⁽⁻¹²⁾ N/C

∴ The Electric Field intensity (E) is 55.73 × 10⁽⁻¹⁵⁾ N/C.

$\rule{300}{1.5}$