Two positive charges of 5.43 µC each are
508 cm apart.
Find the electric field midway between
them.
What is the magnitude of the electric field if
one charge is positive and the other negative,
both of magnitude 5.43 µC?
if they are the same charge, halfway there is zero E.
if they are different,
E=2*kq/(508/2)^2
Why did you do 2 times kq
To find the electric field midway between two positive charges, we can use the formula for the electric field due to a point charge:
Electric field (E) = k * (q/r^2)
Where:
- E is the electric field
- k is the electrostatic constant (k = 9 x 10^9 N m^2/C^2)
- q is the charge of the point charge
- r is the distance from the point charge to the location where the electric field is being measured
Given that the charges are positive and have the same magnitude (5.43 µC), the electric field created by each charge will be the same in magnitude but opposite in direction.
1. Electric field midway between two positive charges:
Since the charges are positive and have the same magnitude, their electric fields will add up at the midpoint. Thus, the electric field midway between them will be the sum of the electric fields due to each charge.
First, convert the distance between the charges from cm to meters:
d = 508 cm = 5.08 m
The electric field due to one charge at the midpoint is given by:
E1 = k * (q/r^2)
E1 = (9 x 10^9 N m^2/C^2) * (5.43 x 10^-6 C) / ((5.08/2)^2)
E1 ≈ 1.26 x 10^5 N/C (rounded to 3 significant figures)
Since the charges have the same magnitude, the electric field due to the second charge will also be approximately 1.26 x 10^5 N/C but in the opposite direction.
The magnitude of the electric field midway between the two charges is the sum of the electric fields due to each charge:
E_mid = E1 + E2
E_mid ≈ (1.26 x 10^5 N/C) + (-1.26 x 10^5 N/C)
E_mid ≈ 0 N/C
Therefore, the magnitude of the electric field midway between the two positive charges is approximately 0 N/C. However, note that the direction of the electric field will depend on the signs of the charges.
2. Magnitude of electric field with one positive and one negative charge:
Since one charge is positive and the other is negative, the electric fields due to each charge will point in opposite directions. The magnitude of both electric fields will be the same since the charges have the same magnitude (5.43 µC).
The electric field due to one charge is given by:
E = k * (q/r^2)
E = (9 x 10^9 N m^2/C^2) * (5.43 x 10^-6 C) / (5.08^2)
E ≈ 6.72 x 10^4 N/C (rounded to 3 significant figures)
Since the charges have opposite signs, the electric fields will point in opposite directions. Therefore, the magnitude of the electric field when one charge is positive and the other is negative is approximately 6.72 x 10^4 N/C.
To find the electric field midway between two positive charges, we can use the formula for electric field:
E = k * (q1 / r1^2) + k * (q2 / r2^2)
Where:
- E is the electric field
- k is the Coulomb constant (8.99 x 10^9 N m^2/C^2)
- q1 and q2 are the magnitudes of the charges (in this case, both 5.43 µC or 5.43 x 10^-6 C)
- r1 and r2 are the distances from the charges to the point where we want to calculate the electric field (in this case, both half of the total distance between the charges, which is 508 cm or 5.08 m)
Now let's calculate the electric field:
E = (8.99 x 10^9 N m^2/C^2) * [(5.43 x 10^-6 C) / (2.54 m)^2] + (8.99 x 10^9 N m^2/C^2) * [(5.43 x 10^-6 C) / (2.54 m)^2]
E = (8.99 x 10^9 N m^2/C^2) * [(5.43 x 10^-6 C) / (2.54 m)^2] + (8.99 x 10^9 N m^2/C^2) * [(5.43 x 10^-6 C) / (2.54 m)^2]
E = 4 * (8.99 x 10^9 N m^2/C^2) * [(5.43 x 10^-6 C) / (2.54 m)^2]
E ≈ 1.7976 x 10^11 N/C
Therefore, the electric field midway between the two positive charges is approximately 1.7976 x 10^11 N/C.
For the second part of the question, where one charge is positive and the other is negative, we can use the same formula. However, since one of the charges is negative, the sign of the electric field will change.
Using the same values as before:
E = (8.99 x 10^9 N m^2/C^2) * [(5.43 x 10^-6 C) / (2.54 m)^2] - (8.99 x 10^9 N m^2/C^2) * [(5.43 x 10^-6 C) / (2.54 m)^2]
E = 0
Therefore, if one charge is positive and the other is negative, both with a magnitude of 5.43 µC, the magnitude of the electric field will be zero at the midpoint between the two charges.