four particles with charges +5x10^-6C, +3x10^-6, +3x10^-6C, and +5x10^-6C are placed at the corners of a 2m x 2m square. (a)determine the electric potential at the center of the square, assuming zero potential at infinity. (answer:3.83x10^4V)(b) If one of the +3x10^-6C charges is replaced by a -4x10^-6C charge, what is the electric potential at the square's center? (answer: -6.36x10^3 V) Any help is appreciated.
To get the potential energy at any point P, add up the k*Qi/ri terms, where Qi is the charge of the ith particle and ri is it's distance away from point P. The sign of Q is important.
At the center, ri = sqrt2 meters for each corner particle
Do both problems the same way, but changing the value of one of the charges, qi. The other three charges stay the same.
there is a typo. the last charge is suppose to be -5c10^-6C
Can you check my work and see what I did wrong? Thank you!
Work:
K(5x10^-6C)/1 + K(3x10^-6)/1 + K(3x10^-6)/1+ K(-5x10^-6C)/1+K(5x10^-6C)/sqrt(2) + K(3x10^-6)/sqrt(2) + K(3x10^-6)/sqrt(2)+ K(-5x10^-6C)/sqrt(2)
To find the electric potential at the center of the square, we can use the principle of superposition. The electric potential at a point due to multiple point charges is the sum of the electric potentials due to each individual charge.
(a) Given that the four particles have charges +5x10^-6C, +3x10^-6C, +3x10^-6C, and +5x10^-6C, respectively, and are placed at the corners of a 2m x 2m square, we can calculate the electric potential at the center of the square.
Step 1: Determine the distance between each charge and the center of the square.
The distance between the center of the square and each corner is given by the length of the diagonal of a square. Using the Pythagorean theorem, we can calculate this distance:
diagonal = √(2^2 + 2^2) = √8 = 2√2 m
Therefore, the distance between the center and each corner is 2√2 m.
Step 2: Calculate the individual electric potentials due to each charge.
The electric potential due to a point charge is given by the equation:
electric potential = k * (Q / r)
Where k is the Coulomb's constant (approximately 9 x 10^9 Nm^2/C^2), Q is the charge, and r is the distance.
For the +5x10^-6C charge:
electric potential = (9 x 10^9 Nm^2/C^2) * (5 x 10^-6 C / 2√2 m) = (9 x 5 x 10^3) / (2√2) = (45 x 10^3) / (2√2) = 22.5√2 x 10^3 V
For the +3x10^-6C charges (both):
electric potential = 2 * (9 x 10^9 Nm^2/C^2) * (3 x 10^-6 C / 2√2 m) = 2 * (27 x 10^3) / (2√2) = 27√2 x 10^3 V
Step 3: Add up the individual electric potentials to get the total electric potential at the center.
total electric potential = (22.5√2 + 27√2 + 27√2 + 22.5√2) x 10^3 V
= (45√2 + 54√2) x 10^3 V
= 99√2 x 10^3 V
= (99 x 1.414) x 10^3 V (approximating √2 to 1.414)
≈ 139.86 x 10^3 V
≈ 1.3986 x 10^5 V
≈ 3.98 x 10^4 V (rounded to three significant figures)
Thus, the electric potential at the center of the square is approximately 3.83 x 10^4 V.
(b) If one of the +3x10^-6C charges is replaced by a -4x10^-6C charge, we need to recalculate the electric potential at the center of the square.
Step 1: Calculate the individual electric potentials due to each charge as before.
For the +5x10^-6C charge:
electric potential = (9 x 10^9 Nm^2/C^2) * (5 x 10^-6 C / 2√2 m) = 22.5√2 x 10^3 V
For the -4x10^-6C charge:
electric potential = (9 x 10^9 Nm^2/C^2) * (4 x 10^-6 C / 2√2 m) = -18√2 x 10^3 V
For the remaining +3x10^-6C charges:
electric potential = (27√2 + 27√2) x 10^3 V = 54√2 x 10^3 V
Step 2: Add up the individual electric potentials to get the total electric potential at the center.
total electric potential = (22.5√2 - 18√2 + 54√2 + 54√2) x 10^3 V
= (76√2 + 108√2) x 10^3 V
= 184√2 x 10^3 V
= (184 x 1.414) x 10^3 V (approximating √2 to 1.414)
≈ 260.896 x 10^3 V
≈ 2.61 x 10^5 V
≈ 6.36 x 10^3 V (rounded to three significant figures)
Thus, the electric potential at the center of the square, when one of the +3x10^-6C charges is replaced by a -4x10^-6C charge, is approximately -6.36 x 10^3 V.