Example 25.4 The Electric Potential Due to a Dipole An electric dipole consists

Question

Example 25.4 The Electric Potential Due to a Dipole An electric dipole consists of two charges of equal magnitude and opposite sign separated by a distance 2aas shown in the figure. The dipole is along the x axis and is centered at the origin.

(A) Calculate the electric potential at point P on the yaxis.

(B) Calculate the electric potential at point R on the +xaxis.

(C) Calculate V and Ex at a point on the x axis far from the dipole. SOLVE IT (A) Calculate the electric potential at point P on the y axis Conceptualize We are seeking the electric potential rather than the electric field.

Categorize Because the dipole consists of only two source charges, the electric potential can be evaluated by summing the potentials due to the individual charges. Find the electric potential at P due to the two charges: VP = ke ? i qi ri = ke ( q ?a2 + y2 + -q ?a2 + y2 ) =
(Use the following as necessary: ke, q, a, and y.)
(A) Calculate the electric potential at point P on the y axis Conceptualize We are seeking the electric potential rather than the electric field.

Categorize Because the dipole consists of only two source charges, the electric potential can be evaluated by summing the potentials due to the individual charges. Find the electric potential at P due to the two charges: VP = ke ? i qi ri = ke ( q ?a2 + y2 + -q ?a2 + y2 ) =
(Use the following as necessary: ke, q, a, and y.)
Find the electric potential at P due to the two charges: VP = ke ? i qi ri = ke ( q ?a2 + y2 + -q ?a2 + y2 ) =
(Use the following as necessary: ke, q, a, and y.)
VP = ke ? i qi ri = ke ( q ?a2 + y2 + -q ?a2 + y2 ) =
(Use the following as necessary: ke, q, a, and y.)
(B) Calculate the electric potential at point R on the +x axis. Find the electric potential at R due to the two charges: VR = ke ? i qi ri = ke ( -q x - a + q x + a ) =
(Use the following as necessary: ke, q, a, and x.)
(B) Calculate the electric potential at point R on the +x axis. Find the electric potential at R due to the two charges: VR = ke ? i qi ri = ke ( -q x - a + q x + a ) =
(Use the following as necessary: ke, q, a, and x.)
Find the electric potential at R due to the two charges: VR = ke ? i qi ri = ke ( -q x - a + q x + a ) =
(Use the following as necessary: ke, q, a, and x.)
VR = ke ? i qi ri = ke ( -q x - a + q x + a ) =
(Use the following as necessary: ke, q, a, and x.)
(C) Calculate V and Ex at a point on the x axis far from the dipole. For point R far from the dipole such that x (C) Calculate V and Ex at a point on the x axis far from the dipole. For point R far from the dipole such that x For point R far from the dipole such that x An electric dipole located on the x axis. An electric dipole consists of two charges of equal magnitude and opposite sign separated by a distance 2aas shown in the figure. The dipole is along the x axis and is centered at the origin. Calculate the electric potential at point P on the yaxis. Calculate the electric potential at point R on the +xaxis. Calculate V and Ex at a point on the x axis far from the dipole. Calculate the electric potential at point P on the y axis Conceptualize We are seeking the electric potential rather than the electric field. Categorize Because the dipole consists of only two source charges, the electric potential can be evaluated by summing the potentials due to the individual charges. Find the electric potential at P due to the two charges: Calculate the electric potential at point R on the +x axis. Find the electric potential at R due to the two charges: Calculate V and Ex at a point on the x axis far from the dipole. For point R far from the dipole such that x >> a, neglect a2 in the denominator of the answer to part (B) and write V in this limit: Calculate the x component of the electric field at a point on the x axis far from the dipole (x >> a): Finalize The potentials in parts (B) and (C) are negative because points on the +x axis are closer to the negative charge than to the positive charge. For the same reason, the x component of the electric field is negative. A charge +q is located at the origin. A charge -2q is at 5.00 m on the x axis. For what finite value of x is the electric field zero? For what finite value(s) of x is the electric potential zero?

Explanation / Answer

A charge +q is located at the origin. A charge -2q is at 5.00 m on the x axis.

(a) For what finite value of x is the electric field zero?

q/x^2 = 2q/(x+5)^2

==> (x+5)/x = sqrt(2)

==> x = -12.1 m

(b) For what finite value(s) of x is the electric potential zero?

q/x = 2q/(x+5)

==> (x+5)/x = 2

==> x = -5 m

q/x = 2q/(5-x)

==> (5-x)/x = 2

==> x = 1.67 m

smaller: -5 m

larger: 1.67 m

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