<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>
An object distance is
presented as s = 5f and we know that the mirror equation relates the image
distance to the object distance and the focal length.
The mirror equation is
1/f = 1/s + 1/s’ where the variable f stands for
the focal length of the mirror. Variable (s)
represents the distance between the mirror surface and the object and the
variable <span>(s’) represents the distance between the mirror surface and
the image. </span>
In addition, a concave mirror
will have a positive focal length (f) and a convex mirror will have a negative
focal length (f).
Now, we then have 1/f = 1/5f
+ 1/s’ which is s’ = 5f/4
Then we get the magnification
ratio that expresses the size or amount of magnification or reduction of the
object or image and to get the magnification, we use this equation: M= s’/s
M= 5f/4x5f
s’ = 1/4s
Therefore, the image height
is one fourth of the object height
Answer:
The photon has a wavelength of 
Explanation:
The speed of a wave can be defined as:
(1)
Where v is the speed, 
is the frequency and 
is the wavelength.
Equation 1 can be expressed in the following way for the case of an electromagnetic wave:
(2)
Where c is the speed of light.
Therefore, 
can be isolated from equation 2 to get the wavelength of the photon.
(3)
Hence, the photon has a wavelength of
<em>Summary: </em>
Photons are the particles that constitutes light.
Choice-C is a correct statement.
Answer:
We see objects in a dark room due to the emission of light photons which are sensitive to our eyes. Darkness is simply a terminology used to describe the absence of light. Visible light to human is a component of the electromagnetic spectrum. Our eyes have receptors that picks the photons which light releases
Explanation:
