for charge +4uC,-2uC,-4uC and +2uC are placed at the corner a,b,c and d of a square of side 20cm calculate net force on+2micro C
Answers
Answer:
Part a)
Net force on +2uC charge is 2.58 N
Part b)
Net force on a fifth positive charge of +5uC placed at the center of the square is 20.12 N
Explanation:
As we know that force between two charges at distance "d" is given as
here force due to corner A is given as
Force due to corner B is given as
Force due to corner C is given as
now we have
Part b)
If another charge is placed at the center of the square then net force on it due to corner A and C
Now for net force due to corner C and D is given as
Now resultant of F1 and F2 is given as
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Topic : electrostatic force
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Answer:
Part a)
Net force on +2uC charge is 2.58 N
Part b)
Net force on a fifth positive charge of +5uC placed at the center of the square is 20.12 N
Explanation:
As we know that force between two charges at distance "d" is given as
F = \frac{kq_1q_2}{d^2}F=
d
2
kq
1
q
2
here force due to corner A is given as
F_1 = \frac{(9\times 10^9)(4\mu C)(2\mu C)}{0.20^2}F
1
=
0.20
2
(9×10
9
)(4μC)(2μC)
F_1 = 1.8 N \hat iF
1
=1.8N
i
^
Force due to corner B is given as
F_2 = \frac{(9\times 10^9)(2\mu C)(2\mu C)}{2(0.20)^2}F
2
=
2(0.20)
2
(9×10
9
)(2μC)(2μC)
F_2 = 0.45(\frac{1}{\sqrt2}(-\hat i - \hat j))F
2
=0.45(
2
1
(−
i
^
−
j
^
))
Force due to corner C is given as
F_3 = \frac{(9\times 10^9)(4\mu C)(2\mu C)}{0.20^2}F
3
=
0.20
2
(9×10
9
)(4μC)(2μC)
F_3 = 1.8 N (-\hat j)F
3
=1.8N(−
j
^
)
now we have
F_{net} = F_1 + F_2 + F_3F
net
=F
1
+F
2
+F
3
F_{net} = 1.8 \hat i + (-0.32 \hat i - 0.32 \hat j) - 1.8 \hat jF
net
=1.8
i
^
+(−0.32
i
^
−0.32
j
^
)−1.8
j
^
F_{net} = 1.48 \hat i - 2.12\hat jF
net
=1.48
i
^
−2.12
j
^
F_{net} = \sqrt{1.48^2 + 2.12^2}F
net
=
1.48
2
+2.12
2
F_{net} = 2.58 NF
net
=2.58N
Part b)
If another charge is placed at the center of the square then net force on it due to corner A and C
F_1 = 2(\frac{(9\times 10^9)(4\mu C)(5\mu C)}{0.5(0.20)^2})F
1
=2(
0.5(0.20)
2
(9×10
9
)(4μC)(5μC)
)
F_1 = 18 NF
1
=18N
Now for net force due to corner C and D is given as
F_2 = 2(\frac{(9\times 10^9)(2\mu C)(5\mu C)}{0.5(0.20)^2})F
2
=2(
0.5(0.20)
2
(9×10
9
)(2μC)(5μC)
)
F_2 = 9 NF
2
=9N
Now resultant of F1 and F2 is given as
F = \sqrt{F_1^2 + F_2^2}F=
F
1
2
+F
2
2
F = \sqrt{9^2 + 18^2}F=
9
2
+18
2
F = 20.12 NF=20.12N