Consider three point charges at the vertices of an equilateral triangle. Let the
ID: 1320048 • Letter: C
Question
Consider three point charges at the vertices of an equilateral triangle. Let the potential be zero at infinity. 60° P 1.99 pC What is the electrostatic potential at the point P at the center of the base of the equilateral triangle given in the dia- gram? The value of the Coulomb constant is 8.98755 x 109 N. ma /C2 Answer in units of V 010 (part 2 of 3) 10.0 points What is the vertical component of the electric force on the 3.9 uC charge due to the 8 pC charge? 1. F (0.48 m) (3.9 pC) (8 C) 2. F (0.48 m2 cot 60° 3. F cos 60° (0.48 2 m) k (3.9 pC) (8 PC) tan 45 4. F (0.48 m)2 kn (3.9 C) (8 yaC) 5. F tan 60 (0.48 m)2 6. F 45 (0.48 m)2 ke (3.9 AC) (8 AuC) 7. F tan 30° (0,48 m) ke (3.9 C) (8 puC) (0.18 m)2 cot 30 9. F cot 45 (0.48 m) 10. F (0.48 m)Explanation / Answer
kq1q2 /r^2 gives the force and not the energy.
Energy means force * distance
hence the( kq1q2 /r^2) * some distance will give energy.
The correct formula for P.E is kq1q2 / r12 where r12 is the distance between the two charges.
kq1q2 /r12 + kq2q3/ r23 + kq1q3/r34
=======================================...
1)
Potential at the point q1 +q2+q3 = -9.91e-6
V = 8.98755e9*(-9.91e-6)/ 0.58 a = -151212/ a
Assuming a = 0.34m
Ptential at the point = -444741.1 V,
==========================
2)
q1q2 +q2q3+q3q1 = (31.2 -15.92 -7.94)* 1.e-12 = -7.4*.e-12
11r2 =r2r3=r3r1 = 0.58 a where a is the side of the triangle
P.E= 8.98755e9*(-32.728*1.e-12)/ 0.58 a
-0.507 a
Assuming a = 0.34m
= -7.4*.e-12 /0.17
P.E = -43.5 J