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Question: 1) What is the force experienced by a test charge of \( 0.20{\text{ }}\mu c \) placed in an electric...

  1. What is the force experienced by a test charge of 0.20 μc0.20{\text{ }}\mu c placed in an electric field of 3.2×1063.2 \times {10^6} NC\dfrac{N}{C}
  2. Define electric field intensity and state its SI unit and dimensions.
  3. Define the following terms for a spherical mirror or surface
    -Pole
    -Centre of curvature
  4. Why does the sky appear blue during the day?
Explanation

Solution

The first question is very simple. You are given the charge q and the electric field; use the formula of force to find the force experienced by the test charge.
F=qEF = qE (Where FF = electrostatic force, qq = charge and EE = electric field)
In the second question, the electric field intensity is a vector quantity and it is the force experienced by the positive charge. The third question is based on the spherical mirrors, where the pole and center of curvature are the part of the mirror. The last question is based on the scattering of light phenomena. Just remember that relation between scattering and wavelength: scattering 1α to wavelengthscattering{\text{ }}\dfrac{1}{\alpha }{\text{ }}to{\text{ }}wavelength .

Complete answer:

  1. In the above question you are given with charge q =0.20 μc= 0.20{\text{ }}\mu c and electric field E=3.2×106E = 3.2 \times {10^6} and you have to find the force. You can directly use the formula of force
    F=qEF = qE (Where FF = electrostatic force, qq = charge and EE = electric field)
    Here charge is given as 0.20 μc0.20{\text{ }}\mu c , you can also write this as 0.2×10−60.2 \times {10^{ - 6}}
    Thus F=qEF = qE
    After substituting the values of qq and EE , the formula becomes:
    ⇒F=0.2×10−6×3.2×106\Rightarrow F = 0.2 \times {10^{ - 6}} \times 3.2 \times {10^6}
    ⇒F=0.2×3.2×10−6+6\Rightarrow F = 0.2 \times 3.2 \times {10^{ - 6 + 6}}
    ⇒F=0.64×100\Rightarrow F = 0.64 \times {10^0}
    ⇒F=0.64 N\Rightarrow F = 0.64{\text{ N}}
    Therefore the force experienced by the test charge =0.64 N= 0.64{\text{ N}}

  2. The electric field intensity- is equal to the force that would be exerted on a small unit positive charge placed as a point.
    The SI unit of electric field intensity is Newton per coulomb and its dimensions are M1L1T−3A−1{M^1}{L^1}{T^{ - 3}}{A^{ - 1}}

  3. In a spherical mirror or surface
    -Pole: the middle point of the reflecting surface of a spherical mirror is called a pole.
    -Centre of curvature: center of curvature of a spherical mirror is the center of the hollow sphere of a glass of which the mirror is a part.

  4. The appearance of the blue color of the sky is due to the phenomenon of scattering of light. You know that the atmosphere has small air molecules, so when sunlight passes through the atmosphere it gets scattered in all directions by these small molecules. You should know that all the colors do not scatter at the same rate. Their scattering depends upon their wavelength. Scattering is inversely proportional to wavelength. Because of this when the white light of the sun passes through these small molecules, blue light having the shortest wavelength scatter most i.e. in all the directions. This is the reason why we see a blue sky on a clear day.

Note:
Remember to convert the unit of charge from microcoulomb to coulomb to answer this question in N/C. Electric field intensity and electric field are different from each other. Do not confuse the two. For a spherical mirror, you can imagine a sphere that has been cut in half. If the silvering portion of the sphere is inside then it is concave if it is outside then it is convex. You should have a clear idea about the scattering of the light phenomenon and also about the visible color and their frequency range.