If Q=20microC and L=60cm, what is the magnitude of the electrostatic force on an
ID: 1259672 • Letter: I
Question
If Q=20microC and L=60cm, what is the magnitude of the electrostatic force on any one of the charges shown? 8. An electron is shot into a homogenous E-field pointing in the Y-direction. At some point in Time, the electron velocity points in the positive X-direction. Given the electric force due to the E-field, which of the following best describes the flight direction of the electron immediately thereafter? a) Between the positive X-axis and the positive Y-axis b) Between the positive X-axis and the negative Y-axis c) Between the negative X-axis and the positive Y-axis d) Between the negative X-axis and the negative Y-axisExplanation / Answer
Electrostatic force between two charged particles q1 and q2 is given by;
F=(1/4pe0)*q1q2/r2
Where, e0 is permittivity of free space.
q1 and q2 are charges.
r=distance between the charges.
The diagonal length is given by;
A=L*sqrt(2)
Now, the electrostatic force on charge +Q would be;
F=(1/4pe0)*(Q)(-Q)/L2+(1/4pe0)*(Q)(Q)/2*L2+(1/4pe0)*(Q)(-Q)/L2
F=(-1/4pe0)*(Q2)/L2+(1/4pe0)* Q2/2*L2-(1/4pe0)* (Q2)/L2
F=(-3/2) (1/4pe0)* (Q2)/L2
As Q=20 mC L=60 cm
Q=20*10^-6 C, L=60*10^-2 m
So,
F=(-3/2) (1/4pe0)* (Q2)/L2
F=(-3/2) (1/4*3.14*8.85*10-12C2 /m2 N)* (20*10^-6 C 2)/( 60*10^-2 m)2
F=7.4*10^5 N