Four point charges qA=2μC, qB=−5μC, qC=2μC, and qD=−5μC are located at the corners of a square ABCD of side 10cm. What is the force on a charge of 1μC placed at the centre of the square?
Four point charges qA=2μC, qB=−5μC, qC=2μC, and qD=−5μC are located at the corners of a square ABCD of side 10cm. What is the force on a charge of 1μC placed at the centre of the square?
Charge placed at the centre is in the influence field of four charges located at the corners of the square. Therefore, we can find force acting on charge placed at the centre using superposition principle. Use the law of vectors to find the net resultant force because force is a vector quantity.
Let the centre of the square is at O. The charge placed on the centre is 1μC.
AB=BC=CD=DA=10cm
AC= √2×10=10√2 cm
AC=BD=10√2cmAO=BO=CO=DO=10√22=5√2cm
Let the force on charge 1μC due to qA is FA which away from both charges qA and q (because both charges are positive in nature, so will repel each other).
The force in charge 1μC due to qc is Fc which away from both qc and q (as they both are positive in nature, so will repel each other).
The force on charge 1μC due to qB is FB which is towards qB (because qB is negatively charged and q is positively charged, so will attract each other).
The force on charge 1μC due to qD is FD which is towards qD (because qD is negatively charged and q is positively charged, so will attract each other).
Force between q and qA
FA=14πε0.qqA(OA)2=9×109×1×10−6×2×10−6(5√2×10−2)2=9×2×10−325×2×10−4FA=9025=185=3.6N (direction towards O to C)
Force between q and qC
FC=14πε0.qqC(OC)2=9×109×1×10−6×2×10−6(5√2×10−2)2=9×2×10−325×2×10−4FC=9025=185=3.6N (direction towards O to A)
Here, we observe that FA and FC are of same magnitude and opposite in direction. So, the resultant force of FA and FC is zero.
Force between q and qB
FB=14πε0.qqB(OB)2=9×109×1×10−6×5×10−6(5√2×10−2)2=9×5×10−325×2×10−4FB=9 N (direction towards O to B)
Force between q and qD
FD=14πε0.qqD(OD)2=9×109×1×10−6×5×10−6(5√2×10−2)2=9×5×10−325×2×10−4FD=9 N (direction towards O to D)
Here, we observe that FB and FD are of some magnitude and opposite in direction. So, the resultant force of FD and FB is zero.
Thus, the net resultant force on 1μC (placed at O) is zero as all the forces balance each other.
Charge placed at the centre is in the influence field of four charges located at the corners of the square. Therefore, we can find force acting on charge placed at the centre using superposition principle. Use the law of vectors to find the net resultant force because force is a vector quantity.
Let the centre of the square is at O. The charge placed on the centre is 1μC.
AB=BC=CD=DA=10cm
AC= √2×10=10√2 cm
AC=BD=10√2cmAO=BO=CO=DO=10√22=5√2cm
Let the force on charge 1μC due to qA is FA which away from both charges qA and q (because both charges are positive in nature, so will repel each other).
The force in charge 1μC due to qc is Fc which away from both qc and q (as they both are positive in nature, so will repel each other).
The force on charge 1μC due to qB is FB which is towards qB (because qB is negatively charged and q is positively charged, so will attract each other).
The force on charge 1μC due to qD is FD which is towards qD (because qD is negatively charged and q is positively charged, so will attract each other).
Force between q and qA
FA=14πε0.qqA(OA)2=9×109×1×10−6×2×10−6(5√2×10−2)2=9×2×10−325×2×10−4FA=9025=185=3.6N (direction towards O to C)
Force between q and qC
FC=14πε0.qqC(OC)2=9×109×1×10−6×2×10−6(5√2×10−2)2=9×2×10−325×2×10−4FC=9025=185=3.6N (direction towards O to A)
Here, we observe that FA and FC are of same magnitude and opposite in direction. So, the resultant force of FA and FC is zero.
Force between q and qB
FB=14πε0.qqB(OB)2=9×109×1×10−6×5×10−6(5√2×10−2)2=9×5×10−325×2×10−4FB=9 N (direction towards O to B)
Force between q and qD
FD=14πε0.qqD(OD)2=9×109×1×10−6×5×10−6(5√2×10−2)2=9×5×10−325×2×10−4FD=9 N (direction towards O to D)
Here, we observe that FB and FD are of some magnitude and opposite in direction. So, the resultant force of FD and FB is zero.
Thus, the net resultant force on 1μC (placed at O) is zero as all the forces balance each other.
