Consider a square of side length a = 9.7 cm with charged particles at each of the corners. Three of the particles carry +9.3 nC of charge while the fourth particle carries -9.3 nC. In volts, what is the electric potential at the center of the square?
Hint: Choose the potential to be zero for a position very far from the square.
Add the voltages from each corner, potential is a scalar, and does not have direction.
V= k(29.3nC)( 2/(.097*.707))
check that.
Isn't it just suppose to be 9.3nc instead of 29.3nC?
To find the electric potential at the center of the square, we can use the principle of superposition. This principle states that the total electric potential at a point due to multiple point charges is the algebraic sum of the electric potentials caused by each individual charge.
In this case, we have four charged particles at the corners of the square. Three particles carry a charge of +9.3 nC, and one particle carries a charge of -9.3 nC.
The electric potential at a point due to a point charge can be calculated using the formula: V = kq/r, where V is the electric potential, k is the Coulomb constant (9 x 10^9 N m^2/C^2), q is the charge of the particle, and r is the distance from the particle to the point where we want to find the potential.
First, let's calculate the electric potential at the center of the square due to each corner particle. Since the square is symmetrical, all four corners are equidistant from the center, so we only need to calculate the potential due to one corner charge.
Using the formula, the electric potential at the center due to one corner particle with a charge of +9.3 nC is:
V1 = (9 x 10^9 N m^2/C^2) * (9.3 nC) / (diagonal distance)
The diagonal distance of the square can be found using the Pythagorean theorem, which gives us:
diagonal distance = √(side length^2 + side length^2) = √(9.7 cm)^2 + (9.7 cm)^2
Calculating this value, we get:
diagonal distance ≈ 13.737 cm
Substituting the values into the equation, we have:
V1 ≈ (9 x 10^9 N m^2/C^2) * (9.3 x 10^-9 C) / (13.737 cm)
Calculating V1, we get:
V1 ≈ 5.865 V
Since there are three corner particles with the same charge, the total electric potential at the center due to these three particles would be: V_total = 3 * V1
V_total ≈ 3 * 5.865 V ≈ 17.595 V
Similarly, the electric potential at the center due to the fourth corner particle with a charge of -9.3 nC would be: V2 = -V1 (opposite sign due to negative charge)
Therefore, the total electric potential at the center of the square would be: V_total = V_total + V2
V_total = 17.595 V + (-5.865 V) = 11.73 V
So, the electric potential at the center of the square is approximately 11.73 volts.