Question

Find the electric intensity of an electric field in which an eletron experiences a force of 2.25N

Answer 1

Answer:

• The Electric Field intensity (E) is 1.40 × 10¹⁹ N/C

Given:

1. Force Experienced (F) = 2.25 N

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 unit:

• 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 = 2.25 N / ( 1.6 × 10⁻¹⁹ C )

∵ [ q = 1.6 × 10⁻¹⁹ C ]

⇒  E = 2.25  / ( 1.6 × 10⁻¹⁹  )

⇒ E = 2.25 × 10¹⁹ / 1.6

⇒ E = 1.40 × 10¹⁹

⇒ E = 1.40 × 10¹⁹ N/C

∴ The Electric Field intensity (E) is 1.40 × 10¹⁹ N/C.

$\rule{300}{1.5}$

Answer 2

AnswEr :

Electric Field Intensity is 14.06 × 10^18 N/C

Explanation :

From the Question,

• Force (F) = 2.25 N

• Charge of an electron (q) = 1.6 × 10^-19 C

To finD

Electric Field Intensity

Electric Field

Intensity of the force experienced by a charge

Mathematically,

$\huge{\star \ \boxed{\boxed{\sf E = \dfrac{F}{q}}}}$

Putting the values,we get :

$\leadsto \sf E = \dfrac{225 \times 10^{-2}}{16 \times 10^{-20}} \\ \\ \leadsto \underline{\boxed{\sf E = 14.06 \times 10^{18} NC^{-1}}}$