a +4C charge is 400m along the horizontal line toward the right of a -3C charge. Calculate the E at a point 300m above the negative charge. ASAP please. It's 7:48pm here and I need it tomorrow at 7am. Thanks.
It is a simple problem, the E vectors add as vectors.
First the E from the negative charge.
E1=kq/r^2=-3Ck/300^2 in the direction of downward.
E2= k(4c)/500^2 in the direction of theta
where Theta=arc tan 300/400 to the left of the upward direction.
E2=4k/500^2 (upward sintheta+left costheta)
E2=4k/500^2 ( 300/500 upward+400/500 to left) check my mental calculations
now add E1+E2
Thanks
To calculate the electric field (E) at a point, you can use Coulomb's Law, which states that the electric field created by a point charge is given by:
E = k * Q / r^2,
where E is the electric field, k is Coulomb's constant (k ≈ 9 × 10^9 N m^2/C^2), Q is the charge of the point charge, and r is the distance from the point charge.
In this case, you have a +4C charge located 400m to the right of a -3C charge, and you need to calculate the electric field at a point 300m above the negative charge.
First, calculate the distance between the point charge and the point where you want to calculate the electric field. This can be found using the Pythagorean theorem since we have a right-angled triangle:
d = sqrt((400m)^2 + (300m)^2)
d = sqrt(160,000m^2 + 90,000m^2)
d ≈ sqrt(250,000m^2)
d ≈ 500m
Now, you can use Coulomb's Law to calculate the electric field:
E = k * Q / r^2
E = (9 × 10^9 N m^2/C^2) * (-3C) / (500m)^2
E = -13500 N/C
Therefore, the electric field at the point 300m above the negative charge is -13500 N/C.
Remember that the negative sign indicates that the electric field points in the opposite direction to the positive charge, which in this case is downwards.