(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.
(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.
<h3>
Magnitude of electric field </h3>
The magnitude of electric field is given by the following equation.
F = qE
But F = mg
mg = qE
E = mg/q
where;
- E is the electric field
- m is mass of the particle
- g is acceleration due to gravity
- q is charge of the particle
<h3>For an electron</h3>
E = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)
E = 5.57 x 10⁻¹¹ N/C
<h3>For proton</h3>
E = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)
E = 1.02 x 10⁻⁷ N/C
Thus, the required vertical electric field is greater when the charge is proton.
Learn more about electric field here: brainly.com/question/14372859
#SPJ1
It’s because flourecent lights operate at higher temperatures than incadecent lights.
When rounding you look at the number right after the one you want to round. The tenth place is the one right after the decimal. If the number after it is more than five it goes up by one, it it's less than five it stays the same. So since the number after the 8 is a 3 it stays the same.
The answer is 7.8
The object represented by this graph is moving toward the origin at constant velocity.
Option 3.
<u>Explanation:</u>
In the figure, x-axis is representing increase in the time and y-axis is presenting increase in the distance from bottom to up. But the line in the graph which is plotted is decreasing from high distance to small distance with increase in time. So this indicates that as the time is increasing, the distance is decreasing.
And the object is moving toward the origin as the distance of the object motion is found to decrease with increase of time as per the graph. But the slope of the graph is found to be almost constant, this indicates that the velocity of the object is constant. Thus, the object represented by this graph is moving toward the origin at constant velocity.
Answer:

Explanation:
The capacitance of a parallel plate capacitor is given by:
(1)
where
is the vacuum permittivity
A is the area of the plates
d is the separation between the plates
The charge stored on the capacitor is given by
(2)
where C is the capacitance and V is the voltage across the capacitor.
The displacement current in the capacitor is given by
(3)
where t is the time elapsed
Substituting (1) and (2) into (3), we find an expression for the displacement current:

where we have



Substituting into the equation, we find
