What is the magnitude of the electric field at a point midway between a −8.3μC and a +5.4μC charge 10cm apart? Assume no other charges are nearby.
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To find the magnitude of the electric field at a point midway between two charges, you can use Coulomb's law.
Coulomb's law states that the magnitude of the electric field, E, created by a point charge is given by the formula:
E = k * (|q1| / r^2)
where k is the electrostatic constant (k = 8.99 x 10^9 Nm^2/C^2), |q1| is the magnitude of the first charge, and r is the distance between the charges.
In this case, the distance between the charges is given as 10 cm, which is equivalent to 0.1 meters. The magnitude of the first charge is 8.3 μC, which is equivalent to 8.3 x 10^-6 C. The magnitude of the second charge is 5.4 μC, which is equivalent to 5.4 x 10^-6 C.
Plugging these values into the formula, we have:
E = (8.99 x 10^9 Nm^2/C^2) * ((8.3 x 10^-6 C) / (0.1 m)^2)
Simplifying:
E = (8.99 x 10^9 Nm^2/C^2) * (8.3 x 10^-6 C / 0.01 m^2)
E = (8.99 x 10^9 Nm^2/C^2) * (8.3 x 10^-6 C / 0.01 m^2)
E ≈ 9.33 x 10^6 N/C
Therefore, the magnitude of the electric field at a point midway between the two charges is approximately 9.33 x 10^6 N/C.
To find the magnitude of the electric field at a point midway between two charges, we can use Coulomb's Law. Coulomb's Law states that the electric field between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula for the electric field (E) is:
E = k * (|q1| + |q2|) / r^2
Where:
E is the magnitude of the electric field
k is the electrostatic constant (9 x 10^9 Nm^2/C^2)
q1 and q2 are the magnitudes (absolute values) of the charges
r is the distance between the charges
In this case, we have q1 = -8.3μC, q2 = +5.4μC, and r = 10 cm.
First, we need to convert the charges to Coulombs (C). Recall that 1 μC = 10^-6 C.
q1 = -8.3μC = -8.3 x 10^-6 C
q2 = +5.4μC = +5.4 x 10^-6 C
Next, we need to convert the distance to meters (m). Recall that 1 cm = 0.01 m.
r = 10 cm = 0.10 m
Now we can substitute the values into the formula:
E = (9 x 10^9 Nm^2/C^2) * ((|-8.3 x 10^-6 C| + |5.4 x 10^-6 C|) / (0.10 m)^2)
Calculating the absolute values of the charges gives us:
E = (9 x 10^9 Nm^2/C^2) * ((8.3 x 10^-6 C + 5.4 x 10^-6 C) / (0.10 m)^2)
Now we solve for E:
E = (9 x 10^9 Nm^2/C^2) * (13.7 x 10^-6 C / (0.10 m)^2)
Simplifying further:
E = (9 x 10^9 Nm^2/C^2) * (137 x 10^-8 C / 0.01 m^2)
E = (9 x 10^9 Nm^2/C^2) * (1.37 x 10^-6 C / 0.01 m^2)
Finally, performing the multiplication:
E = 9 x 1.37 x 10^3 N/C
E = 12,330 N/C
Therefore, the magnitude of the electric field at a point midway between a −8.3μC and a +5.4μC charge 10cm apart is 12,330 N/C.
10 cm = 0.10 meter
so r from each = 0.05
the fields have the same r but add because one is + and the other - and they are in opposite directions so just add charges
Q = (8.8+5.4)*10^-6 coulombs
E = k Q/(0.05)^2