Two point charges qA=-6nC and qB are placed in vacuum. The charge qA is at a distance of 12cm from a point D. The charge qB is at distance 6cm from D. The resultant electric field vector Et created by the two charges at D is along x axis and makes 30 degrees with vector electric field of A and 60 degrees with Vector electric field of B. What is the magnitude of the electric field created by qB at D.
To find the magnitude of the electric field created by qB at point D, we need to use the concept of vector addition of electric fields.
Let's break down the given information:
1. The charge qA is located at a distance of 12 cm from point D.
2. The charge qB is located at a distance of 6 cm from point D.
3. The resultant electric field vector Et created by the two charges at point D is along the x-axis.
4. The resultant electric field vector Et makes an angle of 30 degrees with the electric field vector of charge qA.
5. The resultant electric field vector Et makes an angle of 60 degrees with the electric field vector of charge qB.
First, let's find the magnitudes of the individual electric fields created by qA and qB at point D.
The electric field created by a point charge is given by the equation:
E = k*q / r^2
Where:
- E is the magnitude of the electric field
- k is the electrostatic constant (9 x 10^9 Nm^2/C^2)
- q is the charge of the point charge
- r is the distance between the point charge and the point where the electric field is being calculated
Calculating the magnitude of the electric field created by qA at point D:
EA = k * qA / rA^2
= (9 x 10^9 Nm^2/C^2) * (-6 x 10^-9 C) / (0.12 m)^2
≈ -7.5 x 10^4 N/C (Notice that the negative sign indicates direction)
Calculating the magnitude of the electric field created by qB at point D:
EB = k * qB / rB^2
= (9 x 10^9 Nm^2/C^2) * qB / (0.06 m)^2
Now, let's use the concept of vector addition to find the magnitude of the electric field created by qB at point D (ET) using the given angles:
ET/X = EA/X + EB/X
Since the electric field vector Et is along the x-axis, we can ignore the y-component of the equation.
ET = EA + EB * cos(theta_B)
= (-7.5 x 10^4 N/C) + EB * cos(60 degrees)
We are given that ET makes a 30-degree angle with the electric field vector of charge qA, which means:
ET/Y = EA/Y + EB/Y
ET/Y = EA - EB * cos(theta_B)
= (-7.5 x 10^4 N/C) - EB * cos(60 degrees)
Now, we have two equations:
ET/X = (-7.5 x 10^4 N/C) + EB * cos(60 degrees)
ET/Y = (-7.5 x 10^4 N/C) - EB * cos(60 degrees)
To find EB, we can solve these two equations simultaneously. Let's add them:
ET/X + ET/Y = (-7.5 x 10^4 N/C) + EB * cos(60 degrees) + (-7.5 x 10^4 N/C) - EB * cos(60 degrees)
= (-1.5 x 10^5 N/C)
Now rearranging the equation:
EB = (ET/X + ET/Y) / 2
Substituting the given values:
EB = (-1.5 x 10^5 N/C) / 2
Therefore, the magnitude of the electric field created by qB at point D is approximately -7.5 x 10^4 N/C.