Question

GOAL Use the superposition principle to calculate the electric
field due to two point charges. Consider the following
figure.

The resultant electric field at *P* equals the
vector sum _{1} + _{2}, where _{1} is the
field due to the positive charge *q*_{1}and
_{2} is the field due to the negative charge
*q*_{2}.Two point charges lie along the
*x*-axis in the *x* *y*-coordinate plane.
Positive charge *q*_{1} is at the origin, and
negative charge *q*_{2} is at (0.300 m, 0). Point
*P* is at (0, 0.400 m) on the +*y*-axis. A dashed
line from *q*_{2} to *P* is 0.500 mlong and
makes an acute angle θwith the *x*-axis. Three electric
field vectors start at point *P*.

- Vector
**E**_{1}points straight up in the +*y*-direction. - Vector
**E**_{2}points down and right, at an acute angle θ below the horizontal, along the line from*P*to*q*_{2}. - Resultant vector
**E**points up and right, at an acute angle ϕ above the horizontal.

**PRACTICE IT**

Use the worked example above to help you solve this problem.
Charge *q*_{1} = 6.60 µC is at the origin, and
charge *q*_{2} = -4.90 µC is on the *x*-axis,
0.300 m from the origin (see figure).

**(a)** Find the magnitude and direction of the
electric field at point *P*, which has coordinates (0,
0.400) m.

magnitude | |

direction |

**(b)** Find the force on a charge of
2.00 10^{-8} C placed at *P*.

__________ N

**EXERCISE**

(a) Place a charge of -4.10 µC at point *P* and find the
magnitude and direction of the electric field at the location of
*q*_{2} due to *q*_{1} = 6.60 µC and
the charge at *P*.

magnitude | N/C |

direction | ° counterclockwise from the +x-axis |

(b) Find the magnitude and direction of the force on
*q*_{2}.

magnitude | N |

direction | ° counterclockwise from the +x-axis |

Answer #1

Point charge q1 = -5.00 nC is on the x-axis at x = -0.400 m.
Point P is on the x-axis at x=0.200m. Point charge q2 is at the
origin. What are the sign and magnitude of q2 if the resultant
electric field at point P is zero? Express your answer in
nanocoulombs.

A point charge q2 = - 7.00 nC is located at the origin and
charge q1 = 5.00 nC is on the x-axis, 40.0 cm from the origin.
a) Find the magnitude and direction of electric field at point
P, which has coordinates (40.0 cm,-50.0 cm).
b) Find the force on a charge of 5.00 µC placed at point P.

A point charge q2 = - 7.00 nC is located at the origin and
charge q1 = 5.00 nC is on the x-axis, 40.0 cm from the origin.
a) Find the magnitude and direction of electric field at point
P, which has coordinates (40.0 cm,-50.0 cm).
b) Find the force on a charge of 5.00 µC placed at point P.

A point charge Q1= +5uC is located on the x axis at x = 5.0
cm, and a second point charge Q2= - 7 uC is located on the x axis
at x = -8.0 cm.
1- Calculate the electric field at the origin due to Q1 and
Q2? (give answer in unit vector notations).
2- Calculate the electric field on the y-axis at point y=+6.0
cm due to Q1 and Q2? (give answer in unit vector notations).
3- What...

The total electric field E consists of the vector sum of two
parts. One part has a magnitude of E1 = 1012 N/C and points at an
angle θ1 = 32° above the +x axis. The other part has a magnitude of
E2 = 1667 N/C and points at an angle θ2 = 58° above the +x axis.
Find the magnitude and direction of the total field. Specify the
directional angle relative to the +x axis. magnitude N/C direction
°...

Charge q1 =7.00µc is at origin and charge
q2 =5.00µc is on the X-axis, 0.300m from the origin.
Find the magnitude and direction of the electric field at point p
which has coordinates (0,0.400)m. and find the force on a charge of
2.00*10-3c place at p.

Two charges are placed on the x axis. One of the charges (q1 =
+9.4 µC) is at x1 = +2.7 cm and the other (q2 = -24 µC) is at x2 =
+8.9 cm.
(a) Find the net electric field (magnitude and direction) at x =
0 cm. (Use the sign of your answer to indicate the direction along
the x-axis.)
(b) Find the net electric field (magnitude and direction) at x =
+6.2 cm. (Use the sign of...

1. Find the resultant electric field at the origin, due to Q1
and Q2.
origin = (0,0)
Q1= -70 nC @ (-3m,0m)
Q2= 50 nC @ (4m,-6m)
2. Add a third charge Q3= 25nC @ (2m,8m). Resolve with all 3
charges.

A -3.00 nC point charge is at the origin, and a second -6.50 nC
point charge is on the x-axis at x = 0.800 m.
a. Find the electric field (magnitude and direction) at point on
the x-axis at x = 0.200 m.
b.Find the electric field (magnitude and direction) at point on
the x-axis at x= 1.20 m.
c.Find the electric field (magnitude and direction) at point on
the x-axis at x = -0.200 m.

Two point charges are fixed in place along the x axis
as shown in the figure above. The charge Q1=-3
μμC is located at the origin. The charge Q2=4
μμC is a distance a = 0.46 m to the right of the
origin.
1)
Calculate the x component of the
electric field at point P a distance
h=0.32 m above the origin.
Ex = -0.115 × 106 N/C
Ex = -0.094 × 106 N/C
Ex = 0.264 × 106 N/C...

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