A Charge Q 1 =-16C is fixed on the x-axis at +4.0m, and a charge Q 2 =+9C is fix
ID: 2059802 • Letter: A
Question
A Charge Q1=-16C is fixed on the x-axis at +4.0m, and a charge Q2=+9C is fixed on the y-axis at =+3.0m, as shown.
(a) i. Calculate the magnitude of the electric field E1 at the origin O due to Q1.
ii. Calculate the magnitude of the electric field E2 at the origin O due to Q2.
iii. On the axes to the above and to the left, draw and label vectors to show the electric fields E1 and E2 due to each charge, and also indicate the resultant electric field E at the origin.
(b) Calculate the electric potential V at the origin.
A charge Q3=-4C is brought from a very distant point by an external force and placed at the origin.
(c) On the axes above and to the right, indicate the direction of the force on Q3 at the origin.
(d) Calculate the work that had to be done by the external force to bring Q3 to the origin from the distant point.
Explanation / Answer
a) E1 = k q / r^2 = 8.99x10^9 * 16x10^-6 / 4.0^2 =
= 8990 N/C to the right (because charge is negative... E points toward charge)
E2 = k q / r^2 = 8.99x10^9 * 9x10^-6 / 3.0^2 =
= 8990 N/C downward (because charge is positive... E points away from charge)
At origin, draw E1 to the right and draw E2 down. The resultant is at a 45 deg angle to the right and down (i.e. midway between E1 and E2, because E1 and E2 have the same magnitude. The magnitude of the resultant is 8990 * sqrt2 = 12680 ).
b) elec potential from each charge is just kq/r and since it is a scalar you just add them together, i.e.
total elec potential = 8.99x10^9*(-16x10^-6)/4 + 8.99x10^9*(9x10^-6)/3 =
= -8990 Volts
c) direction of the E field at the origin is down at to the right at 45 deg angle, so the direction of the FORCE on charge Q3 at the origin is up and to the left (i.e. opposite the direction of E) because the charge is negative!
d) work done is the difference in potential energy. At any point, PE = charge * potential and we know that PE is zero at infinity so...
work done = final PE - initial PE = charge * potential at origin - zero =
= -4x10^-6 * -8990 = 0.03596 Joules
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