question archive 1) Draw the electric field lines indicating its direction by arrows for a (a) positive point charge (b) negative point charge (c) electric dipole consisting of a positive and negative charge 2
1) Draw the electric field lines indicating its direction by arrows for a (a) positive point charge (b) negative point charge (c) electric dipole consisting of a positive and negative charge 2
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1) Draw the electric field lines indicating its direction by arrows for a
(a) positive point charge
(b) negative point charge
(c) electric dipole consisting of a positive and negative charge
2. Calculate the magnitude of the E-field produced by a 10 mC point charge at a distance of 0.2 m !
k = 9 x 109 Nm2/C2 Show calculation:
E = _________ N/C
3. Calculate the force acting on a charge of 20 mC placed in an electric field of E = 500 V/m?
Show calculation:
F = _________ N
4. Two positive charges of 4mC each are located at a distance of 10 cm from each other.
What is the magnitude of the force exerted on each charge? Show calculation:
( ) 0.144 N
( ) 1.44 N
( ) 14.4 N
( ) 144 N
( ) other
5. A positive charge is fixed at the origin. A negative charge
is moved along the x-axis from x = 50 cm to x = 10 cm. + - ¬ -
The potential energy of the negative charge
( ) does not change ( ) increases ( ) decreases.
6. A uniform electric field between two parallel plates has a magnitude of 20 N/C and is directed downward.
Draw the E- field indicating with + / - signs
which plate is positive and which is negative?
7. A charge brought into this field experiences a force of 10.0 N downward.
The charge must be (indicate sign of charge with + or -) Show calculation:
( ) +50 C
( ) -50 C
( ) +0.5 C
( ) -0.5 C
8. If the potential at point A is one volt it means that
( ) an object at A has charge equal to one coulomb.
( ) an electron at A can perform one joule of work.
( ) here is a small battery present.
( ) a charge at A has a potential energy of one joule for each coulomb of charge.
9. A capacitor consisting of two parallel plates separated by 4.0 cm has a potential of 100 V on the top
plate and a potential of 0 V on the bottom plate. The value of the electric field in the middle is
( ) 100 N/C Show calculation:
( ) 400 N/C
( ) 2500 N/C
( ) 5000 N/C
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2). E= Kc (Q/r2)
here Q=10*10-6C
r=0.2 meters
substituting the values of k given in the qustion,and Q and r we get the value of electric field strength as =2.25*106 N/C.
3). F=E*Q
E=500,Q=20*10-6 C
Substituting we get F=0.01N
4). F= Kc* (Q1*Q2/r2)
K is known,Q1 and Q2 are same and given as 4 mC , r=0.1M.
substituting for f we get 14.4N............option C
5). DECRESES as the charges accelerate to each other and gains kinetic energy loosing potential energy.
6). figure
7).using same equation as in 3
we use F=10 and E=20
we get +0.5 C positive because it moves towards the negative plate.
8) option D.
9). E=V/d
V=100
d=0.04 m
E=2500 N/C
Please see the attached file for the complete solution
