4. A charge q1 =2.00µC is located at the origin and a charge q2=-6.00µC is located at (0, 3.00) m.
1. Find the total electric potential due to these charges at point P, whose coordinates are (4.00, 0) m.
Vt=k(q1/3+ q2/4)
To find the total electric potential at point P due to charges q1 and q2, we can use the formula for electric potential due to a point charge:
𝑉 = 𝑘 * 𝑞 / 𝑟
Where:
- 𝑉 is the electric potential
- 𝑘 is the electrostatic constant, approximately equal to 9 × 10^9 N⋅m²/C²
- 𝑞 is the charge of the point charge
- 𝑟 is the distance between the point charge and the point of interest
To find the electric potential at point P due to charge q1, we need to calculate the distance between q1 and P, which is the distance between the origin and P:
𝑟₁ = √((4.00 - 0)² + (0 - 0)²)
= √(16.00 + 0)
= 4.00 m
Substituting the values into the formula, we get the electric potential due to q1 at point P:
𝑉₁ = (9 × 10^9 N⋅m²/C²) * (2.00 × 10^-6 C) / (4.00 m)
= (18 × 10^3 N⋅m²/C) / (4.00 m)
= 4.50 × 10^3 N⋅m/C
To find the electric potential at point P due to charge q2, we need to calculate the distance between q2 and P, which is the distance between (0, 3.00) m and (4.00, 0) m:
𝑟₂ = √((4.00 - 0)² + (0 - 3.00)²)
= √(16.00 + 9.00)
= √25.00
= 5.00 m
Substituting the values into the formula, we get the electric potential due to q2 at point P:
𝑉₂ = (9 × 10^9 N⋅m²/C²) * (-6.00 × 10^-6 C) / (5.00 m)
= (-54 × 10^3 N⋅m²/C) / (5.00 m)
= -10.8 × 10^3 N⋅m/C
To find the total electric potential at point P, we sum up the electric potentials due to q1 and q2:
𝑉𝑡𝑜𝑡𝑎𝑙 = 𝑉₁ + 𝑉₂
= 4.50 × 10^3 N⋅m/C + (-10.8 × 10^3 N⋅m/C)
= -6.30 × 10^3 N⋅m/C
Therefore, the total electric potential at point P, whose coordinates are (4.00, 0) m, due to the charges q1 and q2 is approximately -6.30 × 10^3 N⋅m/C.
To find the total electric potential due to the charges at point P, we can use the formula for electric potential:
V = k * (q1 / r1 + q2 / r2)
where V is the electric potential, k is the Coulomb's constant (k = 8.99 x 10^9 Nm^2/C^2), q1 and q2 are the charges, and r1 and r2 are the distances from the charges to point P.
In this case, q1 = 2.00 µC, q2 = -6.00 µC, r1 = the distance from the origin to point P, and r2 = the distance from (0, 3.00) m to point P.
Let's calculate the distances first:
r1 = sqrt((4.00 - 0)^2 + (0 - 0)^2) = sqrt(16) = 4.00 m
r2 = sqrt((4.00 - 0)^2 + (0 - 3.00)^2) = sqrt(25) = 5.00 m
Now we can substitute the values into the formula and solve for V:
V = (8.99 x 10^9 Nm^2/C^2) * ((2.00 x 10^-6 C) / 4.00 m + (-6.00 x 10^-6 C) / 5.00 m)
Simplifying the expression:
V = (8.99 x 10^9 Nm^2/C^2) * (0.5 x 10^-6 C/m - 1.2 x 10^-6 C/m)
V = (8.99 x 10^9 Nm^2/C^2) * (-0.7 x 10^-6 C/m)
V = -6.69 x 10^3 V
Therefore, the total electric potential due to these charges at point P is approximately -6.69 x 10^3 V.