an electric field is vertically acting upwards a small body of mass 1gm and charge -1uC is projected with a velocity 10m/s at an angle 45° with horizontal. its horizontal range is 2m then the intensity of electric field is : (g=10m/s)

To find the intensity of the electric field, we can use the equation for the horizontal range of a projectile in the presence of an electric field.

Given: Mass of the body (m) = 1 gm = 0.001 kg Charge of the body (q) = -1 μC = -1 × 10^-6 C Initial velocity (v) = 10 m/s Launch angle (θ) = 45° Horizontal range (R) = 2 m Acceleration due to gravity (g) = 10 m/s^2

Step 1: Resolve the initial velocity into horizontal and vertical components. The horizontal component of the initial velocity (v_x) can be found using the equation v_x = v × cos(θ). v_x = 10 m/s × cos(45°) = 10 m/s × √2/2 = 5√2 m/s.

Step 2: Calculate the time of flight (t) using the horizontal range and horizontal component of velocity. The time of flight can be found using the equation R = v_x × t. 2 m = 5√2 m/s × t. t = 2 / (5√2) = 2√2 / 10 = √2 / 5 s.

Step 3: Calculate the vertical component of the initial velocity (v_y) using the equation v_y = v × sin(θ). v_y = 10 m/s × sin(45°) = 10 m/s × √2/2 = 5√2 m/s.

Step 4: Calculate the vertical displacement (h) using the vertical component of velocity and time of flight. The vertical displacement can be found using the equation h = v_y × t - (1/2) × g × t^2. h = 5√2 m/s × √2 / 5 s - (1/2) × 10 m/s^2 × (√2 / 5 s)^2. h = 2 m - (1/2) × 10 m/s^2 × 2/5 s^2. h = 2 m - 2 m = 0 m.

Step 5: Calculate the electric field intensity (E) using the vertical displacement and charge of the body. The electric field intensity can be found using the equation E = (m × g) / (q × h). E = (0.001 kg × 10 m/s^2) / (-1 × 10^-6 C × 0 m). E = -0.01 × 10^6 N/C.

Therefore, the intensity of the electric field is -0.01 × 10^6 N/C.