what is the magnitude of the electric field to the nearest tenth of a n/c
1. Calculate the magnitude and direction of the electric field at a point A located at five cm from a signal charge Q = +10 μC. thousand = 9 x x 9 Nm ii C −2 , 1 μC = 10 −six C)
Known :
Electric accuse (Q) = +10 μC = +10 x x -half dozen C
The distance between point A and signal charge Q (r A ) = 5 cm = 0.05 m = 5 ten 10 -2 m
yard = 9 x 10 ix Nm 2 C −2
Wanted: The magnitude and direction of the electric field at indicate A
Solution :
The direction of the electrical field at point A :
The electric charge is positive hence the direction of the electric field away from the electrical charge and points A.
ii. Calculate the magnitude and management of the electric field at a point P located at ten cm from a point charge Q = -two 0 μC. k = 9 ten 10 9 Nm 2 C −2 , 1 μC = 10 −6 C.
Known :
Electrical charge (q) = -20 μC = -20 x ten -6 C
The distance betwixt point P and electric charge (r P ) = x cm = 0.i yard = ane x 10 -1 m
thou = 9 x x 9 Nm two C −2
Wanted: The magnitude and direction of the electric field at point P
Solution :
The management of electrical field at betoken A :
The electrical charge is negative hence the direction of the electric field to the electrical charge.
three. 2 point charges are separated by a distance of 40 cm. What is the magnitude and direction of the electric field at the point P between the two charges, that is 20 cm from indicate A?
Known :
Charge A (q A ) = -2 μC = -2 10 10 -6 C
Charge B (q B ) = + 4 μC = +four x 10 -6 C
The distance between charge A and betoken P (r AP ) = 20 cm = 0.2 m = ii x 10 -1 k
The altitude between charge B and indicate P (r BP ) = 20 cm = 0.2 k = 2 x x -1 k
Wanted : The magnitude and direction of electric field at point P.
Solution :
Charge A is negative so that the management of the electrical field points toward Q A (to the left).
Charge B is positive so that the direction of the electric field points away from Q B (to the left).
The full electric field at point A :
East = E A + E B
E = (4.v ten 10 v ) + (ix x x five )
E = 13.5 x ten 5 N/C
The direction of the electric field points toward Q A (to the left).
4. The magnitude of the electric field is goose egg at…
The charge A is positive and the accuse B is positive and then that the magnitude of the electrical field is nada located at point P, betwixt both charges.
Known :
Charge A (q A ) = + 20 μC = +20 x ten −vi C
Charge B (q B ) = +40 μC = +40 x 10 −half dozen C
k = 9 x 10 9 Nm 2 C −two
The distance between accuse A and the accuse B = 20 cm
The charge betwixt charge A and signal P (r AP ) = a
The distance betwixt charge B and indicate P (r BP ) = 20 – a
Wanted : The magnitude of the electric field is zilch located at….
Solution :
The magnitude of the electric field produced by accuse A at bespeak P
Accuse A is positive so that the direction of the electric field points away from charge A (to the right).
The magnitude of the electric field produced by accuse B at point P :
Charge B is positive so that the management of the electric field points abroad from accuse B (to the left).
The full electric field at point P = 0 :
We utilize the quadratic formula to determine a.
The magnitude of the electric field is zero located at eight cm from accuse A or 12 cm from accuse B.
v . Based on the figure below, west here is the point P so that the electric field at point P is zero? (k = 9 x 10 9 Nm 2 C −2 , one μC = 10 −half dozen C)
Solution
T o calculate the electric field force at point P, causeless at point P there is a positive test charge. Q i is positive and Q 2 is negative, therefore betoken P must be on the correct of Q ii or left of Q 1 . If indicate P is to the left of Q 1 ; the electric field generated by Q ane at the indicate P is to the left (away from Q 1 ) and the electrical field generated by Q 2 at the point P to the right (towards Q 1 ). The direction of the electric field is opposite so that the ii electric fields eliminate each other so that the electrical field strength at point P is zero.
Known :
Q one = +9 μC = +9 x x −half-dozen C
Q 2 = -4 μC = -4 ten ten −6 C
thou = 9 x 10 ix Nm 2 C −two
Distance between charge i and charge 2 = three cm
Distance between Q 1 and point P (r 1P ) = a
Distance between Q two and point P (r 2P ) = three + a
Wanted : location of point P then that the electric field at signal P is zero
Solution :
P oint P is on the left of Q 1 .
The electrical field produced by Q one at point P:
The test charge is positive and Q i is positive and then that the management of an electrical field to leftward.
The electric field produced past Q two at indicate P:
The test charge is positive and Q 2 is negative so that the direction of an electric field to rightward.
Internet electric field at signal A :
Apply the quadratic formula to determine a :
a = -1.25, b = -xiii.five, c = -xx.25
Altitude between Q 2 and point P (r 2P ) = 3 + a = iii – ane.viii = 1.2 cm.
Indicate P is on the 1.2 cm right of Q 1 .
Source: https://physics.gurumuda.net/the-magnitude-and-direction-of-electric-field-problems-and-solutions.htm
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