# Physics 2 question?

Consider the figure shown below. Q1 = 3.00 ×10−9 C and Q2 = -8.80 ×10−9 C. Calculate the potential at the location marked '3'. Charges Q1, Q2, and point '3' are all located at integer coordinates, and assume the potential is zero at infinity.

### 2 Answers

- ArgentLv 72 months agoFavorite Answer
The electric potential Vₑ from a charge Q at a distance r is given by Vₑ = kQ/r, where k is the Coulomb constant: 8.99 × 10⁹ N·m²/C².

The distance r₁ from Q₁ to point 3 is √(2²+4²) = √20 cm ≈ 4.47 cm, or 0.0447 m, or 4.47 × 10ˉ² m.

So the potential at point 3, from Q₁, is

kQ₁/r₁ = (8.99 × 10⁹ N·m²/C²)(3.00 × 10ˉ⁹ C) / (4.47 × 10ˉ² m)

= 6.03 × 10² N·m/C

= 6.03 × 10² J/C

= 6.03 × 10² V, or 603 V.

The distance r₂ from Q₂ to point 3 is √(6²+3²) = √45 cm ≈ 6.708 cm, or 0.06708 m, or 6.708 × 10ˉ² m.

So the potential at point 3, from Q₂, is

kQ₂/r₂ = (8.99 × 10⁹ N·m²/C²)(-8.80 × 10ˉ⁹ C) / (6.708 × 10ˉ² m)

= -1.18 × 10³ V, or -1180 V.

These potentials are scalars, so they simply add algebraically:

603 V + (-1180 V) = -577 V.

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- billrussell42Lv 72 months ago
Q1 = 3.00 ×10−9 C and Q2 = -8.80 ×10−9 C ?? those numbers make no sense

- ArgentLv 72 months agoReport
They make sense if you realize that the "-9" numbers are intended to be exponents: 10ˉ⁹.

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