An insulating sphere of radius 13 cm has a uniform charge density throughout its volume. 13 cm 21.6 cm 7.5 cm p If the magnitude of the elec

Question

An insulating sphere of radius 13 cm has a uniform charge density throughout its volume. 13 cm 21.6 cm 7.5 cm p If the magnitude of the electric field at a distance of 7.5 cm from the center is 78400 N/C , what is the magnitude of the electric field at 21.6 cm from the center? Answer in units of N/C

in progress 0
Trúc Chi 2 days 2021-07-21T22:59:21+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-07-21T23:00:40+00:00

    Answer:

    The value is E_1 = 49224.1 \  N/C

    Explanation:

    From the question we are told that

    The radius is r =  13 \  cm  = 0.13 \  m

    The electric field is E =   78400 \ N/C at a distance d =  7.5 \  cm  =  0.075 \  m

    Generally the electric field at a distance d =  7.5 \  cm  =  0.075 \  m is mathematically represented as

    E =  \frac{k *  q  * d}{r^3} Note: the reason we are using this

    formula is because d < r

    Here k is the coulomb constant with value k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

    => 78400=  \frac{9*10^{9} *  q  * 0.075}{0.13^3 }

    Generally the electric field at a distance d =  21.6  \  cm  =  0.216 \  m is mathematically represented as

    E_1 =  \frac{k *  q  }{^2} Note: the reason we are using this

    formula is because d > r

    Now dividing E_1\ \  by \ \ E

    \frac{E_1}{78400} =  \frac{\frac{9*10^9 * q}{0.216^2} }{\frac{9*10^9 *  q *  0.075}{0.13^3} }

    => E_1 =  \frac{0.13 ^3}{ 0.075 *  0.216^2}  * 78400

    => E_1 = 49224.1 \  J

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )