A +5.7C X 10 to the -6 power point charge is on the x-axis at x=-3.0m,and a +2.0C X 10 to the -6 power point charge is on the x-axis at x=+1.0m.Determine the net electric field (magnitude and direction) on the y-axis at y=+2.0m.
If point P (0, +2) and point A(-3, 0), then PA = sqrt(2^2+3^2) =sqrt(13) = 3.6 m,
If point P (0, +2) and point B(+2, 0), then PB = sqrt(2^2+1^2) =sqrt(5) =2.236 m.
Electric field due to the point charge q1 is
E1=(1/4πε°)• q1/(PA)^2 =9•10^9•5.7•10^-6/13=3946 V/m
An angle PAB=α, sin α =2/3.6 =0.555, cos α = 3/3.6 =0.833
Projections of E1 are
E1(x) =E1•cos α = 3946•0.833= 3287 V/m (directed +x)
E1(y) =E1•sin α = 3946•0.555=2192 V/m (directed +y)
Electric field due to the point charge q2 is
E2=(1/4πε°)• q2/(PB)^2 =9•10^9•2•10^-6/5=3600 V/m
An angle PBA=β, sin β =2/2.236 =0.894, cos β= 1/2.236 =0.447
Projections of E2 are
E2(x) =E2•cos β = 3600•0.894= 3218 V/m (directed -x)
E2(y) =E2•sin β = 3600•0.447=1609 V/m (directed +y)
E(x) = E1(x)-E2(x) =3287 – 3218=69V/m
E(y) = E1(y)+E2(y) =2192 + 1609 = 3801V/m
E =sqrt(E(x)^2+ E(y)^2)) = 3802 V/m
To determine the net electric field at a point on the y-axis, caused by two charges located on the x-axis, we can employ the principle of superposition. This principle states that the total electric field at a point is the vector sum of the electric field due to each individual charge.
In this case, there are two charges on the x-axis. The first charge, +5.7C × 10^-6, is located at x = -3.0m, and the second charge, +2.0C × 10^-6, is located at x = +1.0m. We can calculate the electric field due to each charge separately and then combine them to find the net electric field.
1. Electric field due to the first charge:
The electric field produced by a point charge can be calculated using the formula:
E = (k * q) / r^2
Where:
- E is the electric field magnitude
- k is the Coulomb's constant (k = 9.0 × 10^9 N m²/C²)
- q is the charge magnitude
- r is the distance between the charge and the point where we want to measure the field
In this case, the charge magnitude q1 is +5.7C × 10^-6 and the distance r1 is the y-coordinate of the point (y = +2.0m). So the electric field due to the first charge is:
E1 = (9.0 × 10^9 N m²/C² * 5.7C × 10^-6) / (3.0m)^2
Now, let's perform the calculation to find E1.
2. Electric field due to the second charge:
Similarly, we can calculate the electric field due to the second charge. The charge magnitude q2 is +2.0C × 10^-6, and the distance r2 is also the y-coordinate of the point (y = +2.0m). So the electric field due to the second charge is:
E2 = (9.0 × 10^9 N m²/C² * 2.0C × 10^-6) / (1.0m)^2
Now, let's perform the calculation to find E2.
3. Net electric field on the y-axis:
Finally, to calculate the net electric field on the y-axis, we need to add the magnitudes of E1 and E2. Since they are both on the x-axis, their directions will be opposite. Therefore, the net electric field will be the difference of E1 and E2.
E_net = |E1 - E2|
Now, let's calculate the net electric field magnitude E_net.
After obtaining the magnitudes, we can determine the direction of the net electric field based on the algebraic sign. If E1 is greater than E2, the direction will be towards the negative x-axis. If E2 is greater, the direction will be towards the positive x-axis.
By following these steps, you should be able to determine the net electric field (magnitude and direction) on the y-axis at y=+2.0m.