Question

a point charge q1 = 4.45 nc is located on the x-axis at x = 2.30 m, and a second point charge q2 = -6.80 nc is on the y-axis at y = 1.30 m.

Answer 1

The electric potential at a point P, located in the xy-plane, due to point charges
`\(q_1 = 4.45 \, \text{nC}\) at \(x = 2.30 \, \text{m}\)` and
`\(q_2 = -6.80 \, \text{nC}\) at \(y = 1.30 \, \text{m}\), is given by \(V_P = 5.78 \, \text{V}\).`

Step-by-step explanation:

The electric potential
`\(V_P\)` at point P is the sum of the potentials
`\(V_1\)` and
`\(V_2\)` due to
`\(q_1\)` and
`\(q_2\)` respectively. The potential
`\(V_1\)` at P due to
`\(q_1\)` is given by Coulomb's law:
`\(V_1 = (k \cdot q_1)/(r_1)\),` where
`\(k\)` is Coulomb's constant,
`\(q_1\)` is the charge, and
`\(r_1\)` is the distance from
`\(q_1\)` to P. Similarly,
`\(V_2\)` at P due to
`\(q_2\)` is given by
`\(V_2 = (k \cdot q_2)/(r_2)\)`, where
`\(q_2\)` is the charge, and
`\(r_2\)` is the distance from
`\(q_2\)` to P.

The distances
`\(r_1\) and \(r_2\)` can be calculated using the distance formula
`\(r = √((x - x_0)^2 + (y - y_0)^2)\)`, where
`\(x_0\)` and
`\(y_0\)` are the coordinates of the charges. Substituting the values, we find
`\(r_1 = √((2.30 - 0)^2 + (0 - 0)^2) = 2.30 \, \text{m}\)` and
`\(r_2 = √((0 - 0)^2 + (1.30 - 0)^2) = 1.30 \, \text{m}\).`

Finally, we can find the total potential
`\(V_P\)` by summing
`\(V_1\) and \(V_2\): \(V_P = V_1 + V_2\)`. Substituting the known values, we get
`\(V_P = (k \cdot q_1)/(r_1) + (k \cdot q_2)/(r_2)\)`. Plugging in the numbers, we find
`\(V_P = 5.78 \, \text{V}\).`