Question #311766

1. Two charges are located on the positive x-axis of a coordinate system. Charge 𝑞1 = 2 × 10−9𝐶 is 2cm from the origin, and charge 𝑞2 = −3 × 10−9𝐶 is 4cm from the origin. What is the total force exerted by these two charges on a charge 𝑞3 = 5 × 10−9𝐶 located at the origin?


2. A negative charge of −0.50 × 10−6𝐶 exerts a repulsive force of magnitude 0.20 N on an unknown charge 0.20 m away. What is the unknown charge (magnitude and sign)?


3. How many excess electrons must be placed on each of two small spheres spaced 3cm apart if the spheres are to have equal charge and if the force of repulsion between them is to be 10−19𝑁?


4. What is the electric field 30cm from a charge 𝑞 = 4 × 10−9𝐶?


5. The electric field caused by a certain point charge has a magnitude of 5.0 × 103 𝑁 𝐶 at a distance 0.10m from the charge. What is the magnitude of the charge?





1
Expert's answer
2022-03-15T10:47:10-0400

Fnet=F1+F2F_{net}=F_1+F_2

F=kq1q2r2+kq2q3r2F=\frac{kq_1q_2}{r^2}+\frac{kq_2q_3}{r^2}


F=9×109×5×109(2×1090.022+3×1090.042)=1.40×105NF=9\times10^9\times5\times10^{-9}(\frac{2\times10^{-9}}{0.02^2}+\frac{-3\times10^{-9}}{0.04^2})=1.40\times10^5N

(2)

F=kq1q2r2F=\frac{kq_1q_2}{r^2}

0.20=9×109×0.50×106q0.202q=1.78×106C0.20=\frac{9\times10^9\times0.50\times10^{-6}q}{0.20^2}\\q=1.78\times10^{-6}C

(3)

1019=9×109×q1q20.03210^{-19}=\frac{9\times10^9\times q_1q_2}{0.03^2}

q1q2=1×1032q_1q_2=1\times10^{-32}

q1=q2=qq_1=q_2=q

We know that

q=neq=ne

n2e2=1032n^2e^2=10^{-32}

n=1032(1.6×1019)2=625n=\sqrt{\frac{10^{-32}}{(1.6\times10^{-19})^2}}=625

(4)

Electric field

E=kqr2E=\frac{kq}{r^2}

E=9×109×4×1090.302=4×104N/CE=\frac{9\times10^9\times4\times10^{-9}}{0.30^2}=4\times10^4N/C

(5)Electric field

E=kqr2E=\frac{kq}{r^2}

5×103=9×109×q0.1025\times10^3=\frac{9\times10^9\times q}{0.10^2}

q=0.59×109=q=\frac{0.5}{9\times10^9}=


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