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andriy [413]
3 years ago
14

Two point charges each have a value of 3.0 c and are separated by a distance of 4.0 m. what is the electric field at a point mid

way between the two charges?
Physics
1 answer:
swat323 years ago
4 0
 <span>Place a test charge in the middle. It is 2cm away from each charge. 
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point. 
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out. 
THIS IS A TRICK QUESTION. 
THe electric field exactly midway between them = 0/Q = 0. 
But if the point moves even slightly you need the following formula 
F= (1/4Piε)(Q1Q2/D^2) 
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
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o illustrate the work-energy concept, consider the case of a stone falling from xi to xf under the influence of gravity. Using t
Phantasy [73]

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3 years ago
Assume a beam of light hits the boundary separating medium 1, with index of refraction n1 and medium 2, with index of refraction
Reptile [31]

Answer:

option C

Explanation:

Given,

Refractive index of medium 1 = n₁

Refractive index of medium 2 = n₂

For total internal reflection to take place light should move from denser medium to the rarer medium.

Here Total internal reflection take place at the boundary of medium 1 and medium 2 so, the refractive index of medium 1 is more than medium 2

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3 0
3 years ago
As your skateboard coasts uphill, your speed changes from 3 m/s to 1 m/s in
vlada-n [284]

Answer:

a=-0.33\ m/s^2

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<u>Accelerated Motion</u>

The acceleration of a moving body is defined as the relation of change of speed (or velocity in vector form) with the time taken. The formula is given by

\displaystyle a=\frac{\Delta v}{t}

Or, equivalently

\displaystyle a=\frac{v_f-v_o}{t}

Where vf and vo are the final and initial speeds respectively. The problem gives us these values: v0 = 3 m/s, vf = 1 m/s, t = 3 seconds. Computing a

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A net force of 25N causes an object to accelerate at 4m/s^2. what is the mass of the object?
Doss [256]

Answer:

6.25 kg

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Fnet=ma

m=Fnet/a

m=25/4

m=6.25kg

3 0
3 years ago
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