1. A test charge of +2µC is located 4m to the east of a -5µC charge.(k= 9xNm2/C2) A) Find the electric force felt by the test charge. B) Find the electric field at that location.
2. Calculate the electric field felt by a positive test charge located half way between a charge of +2C and a charge of -2C, that are 3m apart
(1) Q1=+2μCQ2=−5μC(1) \space Q_1=+2\mu C\\Q_2=-5\mu C(1) Q1=+2μCQ2=−5μC
(A) F=kQ1Q2r2=9×109×2×5×10−1216=5.625×10−3 N(A) \space F=\dfrac{kQ_1Q_2}{r^2}=\dfrac{9\times10^9\times2\times5\times10^{-12}}{16}=5.625\times10^{-3}\space N(A) F=r2kQ1Q2=169×109×2×5×10−12=5.625×10−3 N
(B) E=kQ2r2=9×109×5×10−616=2.8125×103 N/C(B)\space E=\dfrac{kQ_2}{r^2}=\dfrac{9\times 10^9\times 5\times 10^{-6}}{16}=2.8125\times10^{3}\space N/C(B) E=r2kQ2=169×109×5×10−6=2.8125×103 N/C
(2) Q1=+2C(2)\space Q_1=+2C(2) Q1=+2C
Q2=−2CQ_2=-2 CQ2=−2C
E1=9×109×2(1.5)2=8×109 N/CE_1=\dfrac{9\times10^9\times2}{(1.5)^2}=8\times10^9\space N/CE1=(1.5)29×109×2=8×109 N/C
E2=9×109×2(1.5)2=8×109 N/CE_2=\dfrac{9\times10^9\times2}{(1.5)^2}=8\times10^9\space N/CE2=(1.5)29×109×2=8×109 N/C
Eres=E1+E2=16×109 N/C(towards left)E_{res}=E_1+E_2=16\times10^9\space N/C(towards\space left)Eres=E1+E2=16×109 N/C(towards left)
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