# Physics: Position and magnitude of electric charge

A +3.0 charge is at and a -1.0 charge is at .

At what point or points on the x-axis is the electric potential zero?

X = ___________ cm

A -2.0 charge and a +2.0 charge are located on the x-axis at and , respectively.

At what position or positions on the x-axis is the electric field zero?

At what position or positions on the x-axis is the electric potential zero?

Estimate the magnitude of the electric field in a cell membrane with a thickness of 8 .

A ion moves from inside a cell, where the electric potential is -70 , to outside the cell, where the potential is 0 .

What is the change in the ion's electric potential energy as it moves from inside to outside the cell?

A 3.0 3.0 parallel-plate capacitor has a 3.0 spacing. The electric field strength inside the capacitor is 1.5×105 .

How much charge is on each plate?

https://brainmass.com/physics/electric-magnetic-fields/physics-position-magnitude-electric-charge-375571

## SOLUTION This solution is **FREE** courtesy of BrainMass!

A +3.0 charge is at and a -1.0 charge is at .

At what point or points on the x-axis is the electric potential zero?

X = ___________ cm

Electric potential at distance r from a point charge q is given by

A) Lat the point is at x = x (x> 4 cm)the potential is zero.

Potential due to positive charge at that point is given by

And field due to negative charge will be

As the net potential at the point will be

Or

Gives 3x - 0.12 = x

Gives x = 0.06 m = 6 cm

B) Lat the point is at x = x (x< 4 cm)the potential is zero.

Potential due to positive charge at that point is given by

And field due to negative charge will be

As the net potential at the point will be

Or

Gives 0.12 -3x = x

Gives x = 0.03 m = 3 cm

Thus the potential is zero at x = 3 cm and x = 6 cm.

A -2.0 charge and a +2.0 charge are located on the x-axis at and , respectively.

At what position or positions on the x-axis is the electric field zero?

At what position or positions on the x-axis is the electric potential zero?

As the two charges are equal and opposite, they constitute an electric dipole of moment P = q*2a where z is the distance from the center (origin here).

The field due to a dipole along a point on its axis (x axis here) is given by

For r > a (on either side of origin) this field will not be zero for any value of x.

For r < a (on either side of origin) the field due to both charges are in same direction and cannot be zero.

Thus the field will not be zero at any point on the x axis except x in infinitely large.

The potential is a scalar quantity and will be zero on the x axis where the two charges are equidistance and thus the potential will be zero at x = 0 itsef.

Thus the potential is zero on the x axis at x = 0. (actually potential is zero at all points on y axis)

Estimate the magnitude of the electric field in a cell membrane with a thickness of 8 .

Electric field also given by potential gradient

E = - dV/dr

Here the potential difference between the cell membrane is near to 80 mV and as the thickness of the cell is 8 nm the order of the field is given by

E = (80*10-3)/(8*10-9)

This is of the order of 107 V/m

Hence the magnitude of electric field in the cell membrane is nearly 107 V/m.

A ion moves from inside a cell, where the electric potential is -70 , to outside the cell, where the potential is 0

What is the change in the ion's electric potential energy as it moves from inside to outside the cell?

The electric potential is the potential energy per unit charge and thus the energy of a charge in a potential field is given by

U = qV

Thus the change of the potential energy of the ion is given by

U = q*V = q(V2 - V1)

Or U = 1.6*10-19[0 - (-70*10-3) = 1.12*10-20 J

Thus the energy of the ion increases by 1.12*10-20 J.

A 3.0 3.0 parallel-plate capacitor has a 3.0 spacing. The electric field strength inside the capacitor is 1.5×105 .

How much charge is on each plate?

The capacitance of a parallel plate capacitor is given by

Here A is the area of the plates and d is the spacing. Thus the capacitance of the capacitor will be

The voltage across the capacitor is given by

V = E*d = 1.5*105*3.0*10-3 = 450 V

Thus the charge on the capacitor plates will be

Q = CV = 2.66*10-12*450 =1.2*10-9 C = 1.2 nC.

The charges on the plates are + 1.2 nC and - 1.2 nC.

https://brainmass.com/physics/electric-magnetic-fields/physics-position-magnitude-electric-charge-375571