An object (with charge -4.2 µC and mass 0.024 kg) hovers at rest above the ground, its weight being held up by a uniform electric field.(a) Find the size and direction of this electric field.(b) If the electric charge on the object is quadrupled while its mass remains the same, find the size and direction of its acceleration.
E must be into the ground, as E is the direction a + test charge would go. A negative charge would to up, which is the case.
Eq=mg
solve for E
then if q is increased by 4x, the net accelerating force is 3E (qQriginal), and that equals m a solve for a. A of course is upwards.
To find the size and direction of the electric field, we need to equate the force due to the electric field to the weight of the object.
(a) Let's start by finding the force due to the electric field (F_el). The force on a charged object in an electric field can be calculated using the formula:
F_el = q * E
where F_el is the force due to the electric field, q is the charge of the object, and E is the electric field strength.
We have the charge of the object as -4.2 µC. However, we need to convert it to coulombs, so we have:
q = -4.2 µC = -4.2 × 10^(-6) C
Now, we equate the force due to the electric field to the weight of the object:
F_el = mg
where m is the mass of the object and g is the acceleration due to gravity.
We have the mass of the object as 0.024 kg. The acceleration due to gravity is approximately 9.8 m/s^2.
Since the object is hovering at rest, the net force on it is zero. Therefore, the electric force must be equal in magnitude but opposite in direction to the weight of the object.
Setting up the equation:
q * E = mg
Plugging in the known values:
(-4.2 × 10^(-6) C) * E = (0.024 kg) * (9.8 m/s^2)
Now we can solve for E:
E = (0.024 kg * 9.8 m/s^2) / (4.2 × 10^(-6) C)
Calculating this, we find:
E ≈ 56,571,428.57 N/C
Since the force due to the electric field is opposite in direction to the weight, the electric field points upwards.
Therefore, the size and direction of the electric field is approximately 56,571,428.57 N/C, pointing upwards.
(b) If the charge on the object is quadrupled while its mass remains the same, the new charge would be:
q_new = 4 * q = 4 * (-4.2 × 10^(-6) C)
Now, let's find the new acceleration of the object using Newton's second law:
F_net = m * a
Since the only force acting on the object is the force due to the electric field, we have:
F_el = m * a
Plugging in the values:
(4 * (-4.2 × 10^(-6) C)) * E = (0.024 kg) * a
Substituting the value of E (previously calculated) and solving for a:
(4 * (-4.2 × 10^(-6) C)) * (56,571,428.57 N/C) = (0.024 kg) * a
We can now solve for a:
a = [(4 * (-4.2 × 10^(-6)) * 56,571,428.57) / 0.024] m/s²
Evaluating this expression, we find:
a ≈ -40,285,714.29 m/s²
The negative sign indicates that the acceleration is in the opposite direction to the electric field.
Therefore, the size and direction of the acceleration is approximately 40,285,714.29 m/s², pointing downwards.