The heart has a dipole charge distribution with a charge of +1.0×107 C that is 5
ID: 1584603 • Letter: T
Question
The heart has a dipole charge distribution with a charge of +1.0×107 C that is 5.0 cm above a charge of -1.0×107 C. Determine the E field (magnitude) caused by the heart's dipole at a distance of 7.8 cm directly above the heart's positive charge. All charges are located in body tissue of dielectric constant 7.0. Determine the E field (direction) caused by the heart's dipole at a distance of 7.8 cm directly above the heart's positive charge. What is the magnitude of the force exerted on a sodium ion (charge +1.6×1019 C) at that point? What is the direction of the force exerted on a sodium ion at that point?
Explanation / Answer
let,
q=1*10^-7C
distance between q1 an q2 is, r=5cm
distance above the q1 charge, d=7.8cm
a)
net electric field, E=E1-E2
=k*q1/d^2 - k*q2/(r+d)^2
=K*q*(1/d^2- 1/(r+d)^2)
=9*10^9*1*10^-7*(1/0.078^2 - 1/(0.05+0.078)^2)
=9.3*10^4 N/C
b)
direction of electric field is upward
c)
F=q*Enet
=1.6*10^-19*(9.3*10^4)
=1.5*10^-14 N
direction of the force is upward
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.