A downward force of 5.1 N is exerted on a -7.3 mcirco C charge. What is the magnitude of the electric field at this point?
E=F/q so E=(5.1N)/(7.3x10^-6 C)
= 6.99x10^5 N/C
7.3 microC = 7x10^-6 C
To find the magnitude of the electric field at a certain point, we can use the equation:
E = F / q
where E is the electric field, F is the force, and q is the charge.
In this case, the force is given as 5.1 N and the charge is given as -7.3 µC (microcoulombs). However, please note that the charge should be in coulombs, so we need to convert it.
1 microcoulomb (µC) = 1 × 10^-6 coulombs (C)
Therefore, -7.3 µC = -7.3 × 10^-6 C
Now, we plug the values into our equation:
E = 5.1 N / (-7.3 × 10^-6 C)
To simplify the calculation, we divide 5.1 N by -7.3 × 10^-6 C:
E = -5.1 N / (7.3 × 10^-6 C)
Calculating this, we get:
E ≈ -698.63 N/C
Since electric field is a vector quantity, the negative sign indicates the direction of the field. In this case, the direction of the field is downward.
Therefore, the magnitude of the electric field at this point is approximately 698.63 N/C, in the downward direction.