An electric field is acting vertically upwards . A particle of mass 1 mg and charge `-1 mu C` is projected with a velocity `20m//s` at an angle `45^(@)` with the horizontal . Its horizontal range is 10 m, then the intensity of electric field is `(g= 10 m//s^(2))`

Options

A. 20, 000 N/C
B. 10,000 N/C
C. 40,000 N/C
D. 90,000N/C

Answer

To solve the problem step by step, we will analyze the motion of the charged particle in the presence of an electric field and calculate the intensity of the electric field based on the given information.

### Step 1: Understand the Forces Acting on the Particle
The particle has a mass \( m = 1 \, \text{mg} = 1 \times 10^{-6} \, \text{kg} \) and a charge \( q = -1 \, \mu\text{C} = -1 \times 10^{-6} \, \text{C} \). The electric field \( E \) is acting vertically upwards, and since the charge is negative, the force due to the electric field \( F_e \) will act downwards.

The gravitational force \( F_g \) acting on the particle is given by:
\[ F_g = mg = (1 \times 10^{-6} \, \text{kg})(10 \, \text{m/s}^2) = 1 \times 10^{-5} \, \text{N} \]

The electric force \( F_e \) is given by:
\[ F_e = qE = (-1 \times 10^{-6} \, \text{C})E = -1 \times 10^{-6}E \, \text{N} \]

### Step 2: Calculate the Net Force and Acceleration
The net force \( F_{\text{net}} \) acting on the particle in the vertical direction is:
\[ F_{\text{net}} = F_g + F_e = 1 \times 10^{-5} - 1 \times 10^{-6}E \]

The net acceleration \( a \) can be expressed as:
\[ a = \frac{F_{\text{net}}}{m} = \frac{1 \times 10^{-5} - 1 \times 10^{-6}E}{1 \times 10^{-6}} = 10 - E \]

### Step 3: Use the Range Formula
The horizontal range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin(2\theta)}{a} \]
where \( u = 20 \, \text{m/s} \) and \( \theta = 45^\circ \). Since \( \sin(90^\circ) = 1 \), we have:
\[ R = \frac{(20)^2}{a} \]
Given that the range \( R = 10 \, \text{m} \), we can set up the equation:
\[ 10 = \frac{400}{10 - E} \]

### Step 4: Solve for Electric Field Intensity \( E \)
Cross-multiplying gives:
\[ 10(10 - E) = 400 \]
Expanding this:
\[ 100 - 10E = 400 \]
Rearranging to find \( E \):
\[ -10E = 400 - 100 \]
\[ -10E = 300 \]
\[ E = -30 \, \text{N/C} \]

Since the electric field is acting upwards and the charge is negative, the force is directed downwards, which is consistent with our calculations.

### Final Answer
The intensity of the electric field is:
\[ E = 30 \, \text{N/C} \]