Question

How much work must I do to assemble a charge distribution consisting of three point charges of -1.00 nC, 2.00nC, and 3.00 nC, located at the vertices of an equilateral triangle? The edges of the triangle are each 20.0 cm in length. Initially the three point charges are infinitely far apart.

Answer 1

`W_(total)=4.5*10^(-8)J`

Step-by-step explanation:

Remember that electric potential can be written as:

`V=(kQ)/(r)`,

Where V is the electric potential, k is Coulomb's constant, Q is a point charge, and r is the distance from the point charge. Also, we can write the electric potential as:

`V=(W)/(q)`,

where W is the work made to move a charge from infinitely far apart to a certain distance, and q the point charge were are moving.

From all this we can get an expression for the work:

`W=(kQq)/(r)`

We are going to let

`q_(1)=-1.00nC\\q_(2)=2.00nC\\q_(3)=3.00nC`

To take the first charge
`q_(1)` from infinitely far apart to one of the vertices of the triangle, since there is no electric field and charges, we make no work.

Next, we will move
`q_(2)` . Now,
`q_(1)` is a vertice of the triangle, and we want
`q_(2)` to be 20.cm apart from
`q_(1 )` so the work we need to do is

`W_(12)=(kq_(1)q_(2))/((0.20))\\\\W_(12)=((9*10^(9))(-1*10^(-9))(2*10^(-9)))/(0.20)\\\\W_(12)=-9*10^(-8)J`

Now, we move the last point charge. Here, we need to take in account the electric potential due to
`q_(1)` and
`q_(2)`. So

`W=W_(13)+W_(23)\\\\\\W=(kq_(1)q_(3))/(0.20)+(kq_(2)q_(3))/(0.20)\\\\W=q_(3)k((q_(1))/(0.20)+(q_(2))/(0.20))\\\\W=(3*10^(-9))(9*10^(9))((-1*10^(-9))/(0.20)+(2*10^(-9))/(0.20))\\\\`

`W=1.35*10^(-7)J`

Now, the only thing left to do is to find the total work, this can be easily done by adding
`W_(12)` and
`W`:

`W_(total)=W_(12)+W\\W_(total)=1.35*10^(-7)-9*10^(-8)\\W_(total)=4.5*10^(-8)J`