Question

Point P is located 58 cm to the right of a fixed point charge of 6.7*10^ -16 C. What is the direction and magnitude of the electric field at point P due to this charge? Let the electrostat constant k = 8.99 * 10 ^ 9 * N * m ^ 2 / (C ^ 2)

Answer 1

The magnitude of the electric field is $1.79\cdot 10^{-5} N/C$, the direction is to the right

Explanation:

The magnitude of the electric field produced by a single point charge is given by

$F=k\frac{q}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q$ is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

$q=6.7\cdot 10^{-16}C$ is the charge

r = 58 cm = 0.58 m is the distance of point P from the charge

Substituting, we find the magnitude of the field at point P:

$E=(8.99\cdot 10^9) \frac{6.7\cdot 10^{-16}}{0.58^2}=1.79\cdot 10^{-5} N/C$

The direction of the field for a positive charge is away from the charge: therefore, since point P is to the right of the charge, the direction of the field is to the right.

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