Three negative charges -1uC,-2uC and -3uC are placed at the corners of equilateral triangle. If the length of each side is 1m, find the magnitude and direction of electric field bisecting the line between the charge q1 and q3.
The electric field at the midpoint between q1 and q3 is given by:
E = (1/4πε_0) * [(q1/r^2) + (q3/r^2)]
where r is the distance between q1 and q3.
In this case, r = 1m, q1 = -1uC, and q3 = -3uC.
Therefore, the electric field at the midpoint between q1 and q3 is:
E = (1/4πε_0) * [(-1uC/1m^2) + (-3uC/1m^2)]
E = (1/4πε_0) * (-4uC/1m^2)
E = -1/(4πε_0) * 4uC/m^2
E = -1N/C
The direction of the electric field is from q3 to q1.
To find the magnitude and direction of the electric field at the point bisecting the line between q1 and q3, we can use the principle of superposition.
Step 1: Calculate the electric field at the point due to charge q1.
The electric field due to a point charge is given by the equation:
E1 = (k * q1) / r1^2
Where:
E1 is the electric field due to q1,
k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2),
q1 is the value of charge q1 (-1uC = -1 x 10^-6 C),
r1 is the distance from q1 to the point bisecting the line.
Since the length of each side of the equilateral triangle is 1m, the distance from q1 to the point bisecting the line is 0.5m (half the length of the side). Plugging in the values:
E1 = (8.99 x 10^9 Nm^2/C^2 * -1 x 10^-6 C) / (0.5m)^2
E1 = -35,960 N/C
Step 2: Calculate the electric field at the point due to charge q3.
Using the same formula as in Step 1, the electric field due to q3 is:
E3 = (k * q3) / r3^2
Where:
E3 is the electric field due to q3,
q3 is the value of charge q3 (-3uC = -3 x 10^-6 C),
r3 is the distance from q3 to the point bisecting the line.
Since the length of each side of the equilateral triangle is 1m, the distance from q3 to the point bisecting the line is 0.5m (half the length of the side). Plugging in the values:
E3 = (8.99 x 10^9 Nm^2/C^2 * -3 x 10^-6 C) / (0.5m)^2
E3 = -107,880 N/C
Step 3: Calculate the net electric field at the point due to charges q1 and q3.
Since the electric field is a vector quantity, we need to consider both the magnitude and direction. To find the net electric field, we add the electric fields due to q1 and q3 using vector addition.
E(net) = E1 + E3
Magnitude of E(net) = |E(net)| = |-35,960 N/C + -107,880 N/C|
Magnitude of E(net) = | -143,840 N/C|
Magnitude of E(net) = 143,840 N/C
The magnitude of the electric field at the point bisecting the line between q1 and q3 is 143,840 N/C.
To determine the direction, we consider the direction of each electric field vector. Since both E1 and E3 are negative and pointing in opposite directions, the net electric field vector will be pointing towards charge q2.
Therefore, the direction of the electric field at the point bisecting the line between q1 and q3 is towards charge q2.
To find the electric field at the midpoint of the line between the charges q1 and q3, we can use the principle of superposition. The electric field due to each charge will be calculated separately, and then we can add them vectorially to find the resultant electric field.
First, let's consider the electric field due to q1 at the midpoint (M) of the line between q1 and q3. Since q1 is negative, the electric field at M will point towards q1. The magnitude of the electric field due to q1 can be calculated using Coulomb's Law:
E1 = k * |q1| / r1^2
where k is the electrostatic constant (9 x 10^9 N.m^2/C^2), |q1| is the magnitude of q1 (-1 x 10^-6 C), and r1 is the distance between q1 and M. In an equilateral triangle, the distance from a vertex to the midpoint of the opposite side is half the length of the side. Therefore, r1 = 1/2 m.
Substituting the values, we get:
E1 = (9 x 10^9 N.m^2/C^2) * (1 x 10^-6 C) / (1/4 m^2)
= 36 x 10^9 N/C
Now, let's move on to the electric field due to q3 at the midpoint (M). Since q3 is also negative, the electric field at M will point towards q3. The magnitude of the electric field due to q3 can be calculated in the same way as E1:
E3 = k * |q3| / r3^2
where |q3| is the magnitude of q3 (-3 x 10^-6 C), and r3 is also equal to 1/2 m.
Substituting the values, we get:
E3 = (9 x 10^9 N.m^2/C^2) * (3 x 10^-6 C) / (1/4 m^2)
= 108 x 10^9 N/C
Now, let's find the resultant electric field at the midpoint (M) by vectorially adding E1 and E3. Since E1 and E3 are pointing in opposite directions, we subtract their magnitudes:
EResultant = E3 - E1
= 108 x 10^9 N/C - 36 x 10^9 N/C
= 72 x 10^9 N/C
Therefore, the magnitude of the electric field at the midpoint of the line between q1 and q3 is 72 x 10^9 N/C. To determine the direction, we note that the resultant electric field is pointing towards q3.
So, the magnitude of the electric field is 72 x 10^9 N/C, and its direction bisects the line between q1 and q3, pointing towards q3.