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Electric Charges and Fields-CBSE Class 12th NCERT Solution

Jul 6, 2013 18:07 IST

    Get, detailed solutions to the questions of the chapter Electric Charges and Fields from NCERT textbooks. The objective is to helping students regarding the pattern of answering the question as per the cbse latest marking scheme.Cbse.jagranjosh.com provided you NCERT solutions for classes 12th math and science subjects.

    Get some questions answered here

    Q. What is the force between two small charged spheres having charges of 2x10-7 C and 3x10-7 C placed 30 cm apart in air?

    Sol.

    Given:
    Q1 = 2x10-7 C
    Q2 = 3x10-7 C
    r = 30x10-2 m
    We know,
    F=(Q_1 Q_2)/(4πε_o r^2 )
    Substituting the given values, we get
    F = 6x10-3 N

    Q. The electrostatic force on a small sphere of charge 0.4 micro coulomb   due to another small sphere of charge -0.8 micro coulomb in air is 0.2 N.

    a) What is the distance between the two spheres?
    b) What is the force on the second sphere due to the first?

    Sol.

    a) From the formula
    F=(Q_1 Q_2)/(4πε_o r^2 )

    Substituting the values, we get
    r = 0.12 m = 12 cm

    b) Force on second sphere due to first
    F21 = (Q_1 Q_2)/(4πε_o r^2 )
    This comes to be -0.2 N, negative sign implying the attractive nature of force. This conforms to Newton’s third law.

    Q. Check that the ratio ke2/Gmemp is dimensionless. Look up a table of Physical Constants and determine the value of this ratio. What does the ratio signify?

    Sol.
    We know very well
                 k = 9x109 Nm2/C2
                 e = 1.6x10-19 C
                 G = 6.6x10-11 Nm2/kg2
                me = 9.11x10-31 kg
                mp = 1.6x10-27 kg

    Ratio = (Nm^2)/C^2 .C^2.〖kg〗^2/〖Nm〗^2 .1/〖kg〗^2  = dimensionless

    And value = 2.4x1039

    The ratio signifies that electrical forces are immensely stronger than the gravitational forces.

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