The number of electrons emitted from the metal per second increases if the intensity of the incident light is increased.
Answer: Option B
<u>Explanation:</u>
As a result of photoelectric effect, electrons are emitted by the light incident on a metal surface. The emitted electrons count and its kinetic energy can measure as the function of light intensity and frequency. Like physicists, at the 20th century beginning, it should be expected that the light wave's energy (its intensity) will be transformed into the kinetic energy of emitted electrons.
In addition, the electrons count emitting from metal must vary with light wave frequency. This frequency relationship was expected because the electric field oscillates due to the light wave and the metal electrons react to different frequencies. In other words, the number of electrons emitted was expected to be frequency dependent and their kinetic energy should be dependent on the intensity (constant wavelength) of light.
Thus, the maximum in kinetic energy of electrons emitted increases with increase in light's frequency and is experimentally independent of light intensity. So, the number of emitted electrons is proportionate to the intensity of the incident light.
Answer:
The answer is below
Explanation:
Given that:
The area of the plates is 6 m by 0.030 m, Therefore the area = 6 m × 0.03 m = 0.18 m²
the relative permittivity of dielectric (εr) is 7.0
Permittivity of free space (εo) = 8.854 × 10^(-12)
capacitance of 100uF
potential difference (V) of 12V
d = separation between plate
The capacitance (C) of a capacitor is given by:
The electric field between plates is given as:
E = V /d
Answer:
The Magnitude of electric field is in the upward direction as shown directly towards the charge 
.
Explanation:
Given:
- side of a square,
- charge on one corner of the square,
- charge on the remaining 3 corners of the square,
<u>Distance of the center from each corners</u>
∴Distance of center from corners, 
Now, electric field due to charges is given as:
<u>For charge we have the field lines emerging out of the charge since it is positively charged:</u>
<u>Force by each of the charges at the remaining corners:</u>
<u> Now, net electric field in the vertical direction:</u>
<u>Now, net electric field in the horizontal direction:</u>
So the Magnitude of electric field is in the upward direction as shown directly towards the charge .
Kinetic energy = 1/2 m v^2 = 1/2 x1.5 x10^-3 x 0.36
