charges of 4uc and -6uc are placed at two corners of an equilateral triangle with sides of 0.1 m .What is the magnitude of the electric field created by these two charges?
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To find the magnitude of the electric field created by two charges placed at the corners of an equilateral triangle, we can follow these steps:
Step 1: Calculate the charge difference (ΔQ)
The charge difference is given by ΔQ = q1 - q2, where q1 = 4 µC and q2 = -6 µC.
ΔQ = 4 µC - (-6 µC)
ΔQ = 4 µC + 6 µC
ΔQ = 10 µC
Step 2: Calculate the distance between the charges (d)
Since the charges are placed at two corners of an equilateral triangle, the side length of the triangle is given as 0.1 m. The distance between the charges is equal to the side length of the triangle.
d = 0.1 m
Step 3: Calculate the electric field magnitude (E)
The magnitude of the electric field is given by the equation E = k * (|ΔQ| / d^2), where k is the electrostatic constant (k = 9 x 10^9 Nm^2/C^2).
Plugging in the values:
E = (9 x 10^9 Nm^2/C^2) * (10 x 10^-6 C / (0.1 m)^2)
E = (9 x 10^9 Nm^2/C^2) * (10 x 10^-6 C / 0.01 m^2)
E = (9 x 10^9 Nm^2/C^2) * (10 x 10^-6 C / 0.001 m^2)
E = (9 x 10^9 Nm^2/C^2) * (10 x 10^-6 C / 1000 m^2)
E = 9 x 10^9 N * 10^-6 / 1000
E = 9 x 10^3 N/C
Therefore, the magnitude of the electric field created by these two charges is 9 x 10^3 N/C.
To find the magnitude of the electric field created by two charges in an equilateral triangle, we can use the principle of superposition.
According to this principle, the electric field at any point is the vector sum of the electric fields created by each individual charge. The electric field due to a point charge is given by Coulomb's law:
E = k * (q / r²),
where E is the electric field, k is the Coulomb's constant (8.99 x 10^9 N⋅m²/C²), q is the charge, and r is the distance between the point charge and the point at which we want to find the electric field.
Let's calculate the electric field at the center of the equilateral triangle, which is the point due to which we want to find the electric field.
Since the sides of the triangle are all equal, we can consider the charges to be at the corners of an equilateral triangle with sides of 0.1 m. Therefore, the distance (r) from each charge to the center of the triangle is also 0.1 m.
The electric field due to the positive charge (4uc) at the center of the triangle is:
E1 = k * (q1 / r₁²) = (8.99 x 10^9 N⋅m²/C²) * (4 x 10^-6 C / (0.1 m)²) = 359.6 N/C.
Since the electric field is a vector quantity, we need to consider its direction. The electric field due to the positive charge points away from the charge. Therefore, at the center of the triangle, this electric field will point radially outwards.
The electric field due to the negative charge (-6uc) at the center of the triangle is:
E2 = k * (q2 / r₂²) = (8.99 x 10^9 N⋅m²/C²) * (-6 x 10^-6 C / (0.1 m)²) = -539.4 N/C.
Similarly, the electric field due to the negative charge points towards the charge. Therefore, at the center of the triangle, this electric field will point radially inwards.
Now, to find the net electric field at the center of the triangle, we need to consider the vector sum of the individual electric fields. Since the two electric fields have the same magnitude but opposite directions, we subtract them:
E_net = E1 + E2 = 359.6 N/C - 539.4 N/C = -179.8 N/C.
Thus, the magnitude of the electric field created by these two charges at the center of the equilateral triangle is 179.8 N/C.