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

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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 =

Physics

1 answer:

Ostrovityanka [942] · Answer 1 · 2026-03-03T01:50:34+03:00

Response:

a) 8.99*10³ V b) 4.5*10⁻² J c) 0 d) 0

Clarification:

a)

Electrostatic potential V is defined as the work achieved per unit charge, as conducted by the electrostatic force, which moves a distance d from infinity (considered the reference zero level).
For a point charge, it can be represented mathematically:

$V =\frac{k*q}{d}$

Since electrostatic force behaves linearly concerning charge, we can apply the superposition principle.
This principle states that the cumulative potential at any given point is simply the sum of the individual potentials contributed by various charges, as if the others were absent.
In our specific configuration, due to symmetry, the potential at each corner of the triangle is simply double that of the potential resulting from any charge at another corner, as demonstrated:

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

The potential at point C registers as 8.99*10³ V.

b)

The energy needed to move a positive charge of 5μC from infinity to point C is calculated by multiplying the potential at that point by the charge, explained below:

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

The work required amounts to 0.045 J.

c)

If we substitute one of the charges at point C with one of the opposite charge of equal magnitude, the following equation emerges:

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

This indicates that the potential arising from both charges results in 0 at point C.

d)

With point C's potential calculated as 0 and assuming V=0 at infinity too, we derive that bringing the charge of 5μC from infinity to point C requires no work, as there is no potential difference between the two locations.