A uniform electric field of 2.1 x 106 N/C is directed in the positive x direction. A charge of -4.3 micro-Coulombs with a mass of 7.5 kg is initially moving at 18 m/s in the positive x direction before it enters the field. How many seconds does it take before the charge stops?
To find the time it takes for the charge to stop, we can use the equation of motion for uniformly accelerated motion.
The equation is given as: v = u + at
Where:
v = final velocity (0 m/s, since the charge stops)
u = initial velocity (18 m/s in the positive x direction)
a = acceleration (we'll calculate this using the electric field)
t = time (what we need to find)
First, let's find the acceleration of the charge using the force exerted by the electric field.
The force experienced by the charge in an electric field is given by the equation: F = qE
Where:
F = force acting on the charge
q = charge of the particle (-4.3 micro-Coulombs, which is equivalent to -4.3 x 10^-6 Coulombs)
E = electric field strength (2.1 x 10^6 N/C)
Plugging in the values, we get the force as:
F = (-4.3 x 10^-6 C) * (2.1 x 10^6 N/C)
F = -9.03 N (since the charge is negative, the force is also negative)
Now, we know that force (F) is equal to mass (m) multiplied by acceleration (a), so we can rearrange the equation to solve for a as:
a = F / m
Plugging in the values, we get:
a = (-9.03 N) / (7.5 kg)
a = -1.204 m/s^2 (again, negative since the force is negative)
Now, let's go back to the equation of motion:
v = u + at
Since the charge stops (v = 0), we can rearrange the equation to solve for time (t):
t = (v - u) / a
Plugging in the values, we get:
t = (0 - 18 m/s) / (-1.204 m/s^2)
Calculating this, we get:
t ≈ 14.940 seconds
Therefore, it takes approximately 14.940 seconds for the charge to stop in the given uniform electric field.