Question

Points A, B, and C are at the corners of an equilateral triangle of side 8 m. Equal positive charges of 4 mu or micro CC are at A and B. (a) What is the potential at point C? 8.990 kV * [2.5 points] 2 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 8.990 OK (b) How much work is required to bring a positive charge of 5 mu or micro CC from infinity to point C if the other charges are held fixed? .04495 J * [2.5 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] .04495 OK (c) Answer parts (a) and (b) if the charge at B is replaced by a charge of -4 mu or micro CC. Vc= kV [2.5 points] 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] W =

Answer 1

Answer:

a) 8.99*10³ V  b) 4.5*10⁻² J c) 0 d) 0

Explanation:

a)

• The electrostatic potential V, is the work done per unit charge, by the electrostatic force, producing a displacement d from infinity (assumed to be the reference zero level).
• For a point charge, it can be expressed as follows:

        $V =\frac{k*q}{d}$

• As the electrostatic force is linear with the charge (it is raised to first power), we can apply superposition principle.
• This means that the total potential at a given point, is just the sum of the individual potentials due to the different charges, as if the others were not there.
• In our case, due to symmetry, the potential, at any corner of the triangle, is just the double of the potential due to the charge located at  any other corner, as follows:

        $V = \frac{2*q*k}{d} = \frac{2*8.99e9N*m2/C2*4e-6C}{8m} =\\ \\ V= 8.99e3 V$

• The potential at point C is 8.99*10³ V

b)

• The work required to bring a positive charge of 5μC from infinity to the point C, is just the product of the potential at this point times the charge, as follows:

        $W = V * q = 8.99e3 V* 5e-6C = 4.5e-2 J$

• The work needed is 0.045 J.

c)

• If we replace one of the charges creating the potential at the point  C, by one of the same magnitude, but opposite sign, we will have the following equation:

       $V = \frac{8.99e9N*m2/C2*(4e-6C)}{8m} + (\frac{8.99e9N*m2/C2*(-4e-6C)}{8m}) = 0$

• This means that the potential due to both charges is 0, at point C.

d)

• If the potential at point C is 0, assuming that at infinity V=0 also, we conclude that there is no work required to bring the charge of 5μC from infinity to the point C, as no potential difference exists between both points.