Let the midpoint of the two charges be considered as the origin, and charge A = 30.0 * 10⁻⁹ C be moving in the +x- axis and the charge B = 60.0 * 10⁻⁹ C be moving in the -x-axis.

Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹ N.m²/C², Q = quantity of charge, r = distance of separation

In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m

Electric field due to charge A

Ea = +(9.0 * 10⁹ N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²

Ea = +1.8 * 10⁴ N/C

Electric field due to charge B

Eb = -(9.0 * 10⁹ N.m²/C² * 60.0 * 10⁻⁹ ) / ( 0.015 m)²

Eb = -3.6 * 10⁴ N/C

The resultant electric field E = Ea + Eb

E = (+1.8 * 10⁴ + -3.6 * 10⁴) N/C

E = -1.8 * 10⁴ N/C

Therefore, the electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

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3 years ago

How could professional football players use technology to better understand their health

stealth61 [152]

Athletes of any kind can use technology to monitor things like their BMI (body mass index) or weight or anything of the sort to target what they need to work on to improve their health.

8 0

3 years ago

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Unpolarized light with an average intensity of 845 W/m2 enters a polarizer with a vertical transmission axis. The transmitted li

RideAnS [48]

The concept to develop this problem is the Law of Malus. Which describes what happens with the light intensity once it passes through a polarized material.

Mathematically this can be expressed as

$I = I_0 cos^2\theta$

Where

I = New intensity after pass through the Polarizer

$I_0$ = Original intensity

$\theta$ = Indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

When the light passes perpendicularly through the first polarizer, the light intensity is reduced by half which will cause the intensity to be $225W / m ^ 2$ at the output of the new polarizer, mathematically: