Answer:
- 1.3 x 10⁻¹⁵ C/m
Explanation:
Q = Total charge on the circular arc = - 353 e = - 353 (1.6 x 10⁻¹⁹) C = - 564.8 x 10⁻¹⁹ C
r = Radius of the arc = 5.30 cm = 0.053 m
θ = Angle subtended by the arc = 48° deg = 48 x 0.0175 rad = 0.84 rad (Since 1 deg = 0.0175 rad)
L = length of the arc
length of the arc is given as
L = r θ
L = (0.053) (0.84)
L = 0.045 m
λ = Linear charge density
Linear charge density is given as
Inserting the values
λ = - 1.3 x 10⁻¹⁵ C/m
Answer:
Explanation:
We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.
1 m =100 cm
Surface area =S=
We have to find the charge Q on the positive plates of the capacitor.
V=Initial voltage between plates
d=Initial distance between plates
Initial Capacitance of capacitor
Capacitance of capacitor after moving plates
Potential difference between plates after moving
Hence, the charge on positive plate of capacitor=
Answer:
Acceleration of the proton will be equal to
Explanation:
We have given electric field
Mass of proton is equal to
And charge on proton is equal to
Electrostatic force will be responsible for the motion of proton
Electrostatic force will be equal to
According to newton law force on the proton will be equal to F = ma, here m is mass of proton and a is acceleration
This newton force will be equal to electrostatic force
So
So acceleration of the proton will be equal to
Answer:
D. 3 psi
Explanation:
Pressure is defined as force acting per unit area and is numerically expressed as:
where P represent pressure, F is force and A is area where the force acts. Substituting 300 lb for force and 100 sq.in for area then pressure,
Therefore, from the choices given, option D, 3psi is the right choice.