The heart has a dipole charge distribution with a charge of +1.0×107 C that is 5

Question

The heart has a dipole charge distribution with a charge of +1.0×107 C that is 5.3 cm above a charge of -1.0×107 C.

Part A

Determine the E  field (magnitude) caused by the heart's dipole at a distance of 8.2 cm directly above the heart's positive charge. All charges are located in body tissue of dielectric constant 7.0.

Part B

Determine the E  field (direction) caused by the heart's dipole at a distance of 8.2 cm directly above the heart's positive charge.

Part C

What is the magnitude of the force exerted on a sodium ion (charge +1.6×1019 C) at that point?

Part D

What is the direction of the force exerted on a sodium ion at that point?

Explanation / Answer

electric field due to + charge

E+ = q/(4*pi*k*eo*r^2) = (10^-7)/(4*pi*7*8.84*10^-12*0.082^2) = +133848.89 N/C

E+ = 19125.5 N/C

electric field due to - charge

E_ = q/(4*pi*k*eo*r^2) = (10^-7)/(4*pi*7*8.84*10^-12*(0.082+0.053)^2) = 133848.89 N/C

E_ = -7056.223 N/C

E = E+ + E_

E = 26181.723 N/C

++++++++++++++++

B)

direction upwards

(c)

F = Eq = 26181.723*1.6*10^-19 = 4.18907568*10^-15 N

D)

direction up wards

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