A. Find the electric potential energy for the following array of charges: charge

Question

A. Find the electric potential energy for the following array of charges: charge q1 = +4.0 µC is located at (x, y) = (0.0, 0.0) m; charge q2 = +2.0 µC is located at (4.0, 3.0) m; and charge q3 = -1.0 µC is located at (0.0, 3.0) m. answer in mJ

B. How much work is done by an applied force that moves two charges of +7.0 µC that are initially very far apart to a distance of 3.9 cm apart? in joules

C. If an electron moves from one point at a potential of -100.0 V to another point at a potential of +100.0 V, how much work is done by the electric field?

Explanation / Answer

A) charge q1 =+4 C at (0 m, 0m) charge q2 =+2 C at (4 m, 3 m) charge q3 =-1 C at (0 m, 3 m) r12 =(4-0)2+(3-0)2 =5 m similarly, r13 =3 m   and r23 =4 m electric potential energy U =kqq'/r ......... (1) from eq (1), electric potential energy due to charge q1 and q2 is U12 = kq1q2/r12   ......... (2) where, k =8.99*10^9 N.m^2/C^2 electric potential energy due to charge q1 and q3 is U13 = kq1q3/r13   ......... (3) electric potential energy due to charge q2 and q3 is U23 = kq2q3/r23   ......... (4) substitute the given data in eq (2), we get U12 =14.38*10^-3 J =14.38 mJ substitute the given data in eq (3), we get U13 =-11.98 mJ substitute the given data in eq (4), we get U23 =-4.5 mJ therefore, total electric potential energy is U =U12+U13+U23 =14.38-11.98-4.5 U=-2.095 mJ ......................................................... B) charge q1 =q2 =q =+7 C =+7*10^-6 C r =3.9 cm =3.9*10^-2 m work done W =kq1q2/r    ........... (5)                  W =11.3 J ................................................................ C) Va =-100 V and Vb =100 V charge of electron q =1.6*10^-19 C work done W =U =q(Vb-Va)                             =3.2*10^-17 J electric potential energy due to charge q1 and q3 is U13 = kq1q3/r13   ......... (3) electric potential energy due to charge q2 and q3 is U23 = kq2q3/r23   ......... (4) substitute the given data in eq (2), we get U12 =14.38*10^-3 J =14.38 mJ substitute the given data in eq (3), we get U13 =-11.98 mJ substitute the given data in eq (4), we get U23 =-4.5 mJ therefore, total electric potential energy is U =U12+U13+U23 =14.38-11.98-4.5 U=-2.095 mJ ......................................................... B) charge q1 =q2 =q =+7 C =+7*10^-6 C r =3.9 cm =3.9*10^-2 m work done W =kq1q2/r    ........... (5)                  W =11.3 J ................................................................ C) Va =-100 V and Vb =100 V charge of electron q =1.6*10^-19 C work done W =U =q(Vb-Va)                             =3.2*10^-17 J electric potential energy due to charge q2 and q3 is U23 = kq2q3/r23   ......... (4) substitute the given data in eq (2), we get U12 =14.38*10^-3 J =14.38 mJ substitute the given data in eq (3), we get U13 =-11.98 mJ substitute the given data in eq (4), we get substitute the given data in eq (4), we get U23 =-4.5 mJ therefore, total electric potential energy is U =U12+U13+U23 =14.38-11.98-4.5 U=-2.095 mJ ......................................................... B) charge q1 =q2 =q =+7 C =+7*10^-6 C r =3.9 cm =3.9*10^-2 m work done W =kq1q2/r    ........... (5)                  W =11.3 J ................................................................ C) Va =-100 V and Vb =100 V charge of electron q =1.6*10^-19 C work done W =U =q(Vb-Va)                             =3.2*10^-17 J

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