A point charge Q of -9.86 C and a large flat plate with surface charge density 2.15 C/m2 together create an electric field at point P. Charge Q and point P are each a distance a = 5.03 m from the plate. Point P is a distanceb = 9.25 m from charge Q. Find the magnitude of the force exerted by the electric field on a particle with charge 4.54 nC placed at point P in the figure. Answer in units of N. plate (side view)

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A point charge Q of -9.86 C and a large flat plate with surface charge density 2.15 C/m2 together create an electric field
5.03 m from the plate. Point P is a distance b = 9.25 m from
at point P. Charge Q and point P are each a distance a =
charge Q. Find the magnitude of the force exerted by the electric field on a particle with charge 4.54 nC placed at point
P in the figure. Answer in units of N.
plate (side view)
Transcribed Image Text:A point charge Q of -9.86 C and a large flat plate with surface charge density 2.15 C/m2 together create an electric field 5.03 m from the plate. Point P is a distance b = 9.25 m from at point P. Charge Q and point P are each a distance a = charge Q. Find the magnitude of the force exerted by the electric field on a particle with charge 4.54 nC placed at point P in the figure. Answer in units of N. plate (side view)
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Given

Given:

Charge,q=-9.86nC=-9.86×10-9CSurface Charge Density,σ=2.15C/m2Distance betwen the charge Q and point P,a=5.03mDistance of point P,b=9.25mCharge,q2=4.54nC=4.54×10-9CPermitivity of free space,ε0=8.85×10-12C2/Nm2Coulomb Constant,k=9×109Nm2/C2

The electric force per unit charge is defined as the electric field. The force that the field would exert on a positive test charge is assumed to be in the same direction as the field's direction. A positive charge's electric field is radially outward, whereas a negative charge's field is radially inside.

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