# Calculate the electric field intensity at a point 3 cm away from point charge of 3 x 10. show your work

###### Question:

show your work

## Answers

The intensity of the electric field is 30,000 N/C

Explanation:

The strength of the electric field produced by a single-point charge is given by the equation

[tex]E=k\frac{q}{r^2}[/tex]

where:

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

q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

[tex]q=3\cdot 10^{-9}C[/tex] is the magnitude of the charge

r = 3 cm = 0.03 m is the distance at which we are calculating the field intensity

Substituting, we find:

[tex]E=(8.99\cdot 10^9)\frac{3\cdot 10^{-9}}{(0.03)^2}=30,000 N/C[/tex]

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3*10^14v/m

Explanation:

[tex]calculate the electric field intensity at a point 3 cm away from point charge of 3 x 10. show your w[/tex]

The electric field intensity is 30000 N/C.

Explanation:

Given:

Magnitude of the point charge is, [tex]q=3\times 10^{-9}\ C[/tex]

Distance of the given point from the point charge is, [tex]d=3\ cm=0.03\ m[/tex]

Electric field intensity is directly proportional to the magnitude of point charge and inversely proportional to the square of the distance of the point and the given charge.

Therefore, electric field intensity 'E' at a distance of 'd' from a point charge 'q' is given as:

[tex]E=\frac{kq}{d^2}[/tex]

Plug in [tex]k=9\times 10^9\ N\cdot m^2/C^2, q=3\times 10^{-9}\ C, d=0.03\ m[/tex]. Solve for 'E'.

[tex]E=\frac{(9\times 10^9\ N\cdot m^2/C^2)(3\times 10^{-9}\ C)}{(0.03\ m)^2}\\\\E=\frac{27}{0.0009}\ N/C\\\\E=30000\ N/C[/tex]

Therefore, the electric field intensity at a point 3 cm from the point charge is 30000 N/C.

The intensity of the electric field is 30 000 N/C

Explanation:

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

[tex]F=k\frac{q}{r^2}[/tex]

where:

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

[tex]q[/tex] is the magnitude of the charge

r is distance from the charge

In this problem, we have:

[tex]q=3\cdot 10^{-9}C[/tex] is the magnitude of the charge

r = 3 cm = 0.03 m is the distance from the charge

Substituting into the equation, we find the intensity of the field:

[tex]E=(8.99\cdot 10^9) \frac{3\cdot 10^{-9}}{0.03^2}=30,000 N/C[/tex]

Learn more about electric fields:

#LearnwithBrainly