Find the force (in N) on a -4.43 mu C point charge from an external electric fie

Question

Find the force (in N) on a -4.43 mu C point charge from an external electric field of 555 kN/C toward the top Vector Equation used Solution (in vector form) F_0 = (-4.43 times 10 C) (555 times 10 N/C) The electron and the proton are 52.9 pm apart in the plane of the paper. Find their electric dipole moment The one equation used (NOT p = qd) SOLUTION electron 52.9 pm proton The electric dipole moment of an electric dipole is 8.47 time 10^-30 C middot m right (rightarrow) in an external electric field of 5.55 times 10^5 N/C toward the top (uparrow). Find the torque this external electric field exerts on the electric dipole. VECTOR EQUATION USED SOLUTION (OF THE MAGNITUDE) I used phi = in finding the magnitude as follows: |torque| = phi(symbols), so numerically |torque| = Find the electric field that the proton sets up at point P (the position of the electron in Problem 2 above) The one equation used p 52.9 pm proton Two equal negative point charges are on the x-axis in vacuum at x = -7.00 mm and +2.00 mm as shown. The magnitude of their Coulomb's law force is 7.12 times 10^-23 N. Find the charge of either one and the direction of force the electric field of charge 1 exerts on charge 2. The one equation used solution 1 2

Explanation / Answer

2)

Electric dipole moment, P = q*d

= 1.6*10^-19*52.9*10^-12

= 8.46*10^-30 C.m

direction = towards proton (right)

3)

Torque = E*d = 5.55*10^5*52.9*10^-12

= 2.94*10^-5 N.m <--- magnitude

direction = out of the page

4)

E = k*q/d^2

= 9*10^9*1.6*10^-19/(52.9*10^-12)^2 = 5.15*10^11 N

direction = towards left

5)

F = k*q^2/d^2

So, 7.12*10^-23 = 9*10^9*(q^2)/((7+2)*10^-3)^2

So, q = 8*10^-19 C

direction = towards left

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