Point P, is where the electric field is zero

Given are two charges, q1 = 40 nC and q2 = -20 nC, separated by 14 cm.

Firstly, we find the distance x from the 40 nC charge where the electric field is zero. This is where the magnitudes of the electric fields produced by the two charges are equal, i.e.,

|E1| = |E2|.

Coulomb's law gives the electric field E produced by a charge q at a distance x:

E = kq/x², where k is Coulomb's constant.

Setting |E₁| = |E₂| gives us:

|kq₁ / (x)²| = |kq₂ /(x - 14cm)²|

Solving this equation for x gives us the point P:

q₁ / (x)² = q₂ / (x - 14cm)²

40.0 nC / (x)² = -20.0nC / (x - 14cm)²

Simplifying

2.0 / (x)² = -1.0 / (x - 14cm)²

2.0 (x - 14cm)² = -1.0(x)²

x² - 56x + 392 =0

(x - 49)(x - 7) = 0

x = 49 cm = x = 7.0cm

Now we can calculate the electric potential V at point P. The electric potential V produced by a charge q at a distance r is given by V = k q / r

So the total electric potential at point P due to both charges is given by:

V = V₁ + V₂ = k[{q₁/(r₁)} + {q₂/(r₂)}]

V = k[{q₁/(r₁)} + {q₂/(r₂)}]

V = (9.0 × 10⁹ Nm²/C²) [{40 nC/(0.49 m)} + {-20nC /(0.49 m - 0.14cm)}]

V = +220 volt

Therefore, the electric potential at point P, where the electric field is zero, is +220V.

A uniform electric field with strength 12 V/m points from point A to point B, which are on the x-axis: xA=45⁢m and xB=12⁢m. What is the potential difference VA−VB ?

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A uniform electric field with strength $12 V / m$ points from point A to point B, which are on the x-axis: $x_{A} = 45 m$ and $x_{B} = 12 m$ . What is the potential difference $V_{A} - V_{B}$ ?