Answer:
The potential energy at point 'P' due to the first charge can be found as follows:
V1=kq1r1V1=(9.0×109N.m2/C2)(+1.0×10−5C)1.0mV1=9.0×104VV1=kq1r1V1=(9.0×109N.m2/C2)(+1.0×10−5C)1.0mV1=9.0×104V
The potential energy at point 'P' due to the second charge can be found as follows:
V2=kq2r2V2=(9.0×109N.m2/C2)(−5.0×10−6C)1.0mV2=−4.5×104VV2=kq2r2V2=(9.0×109N.m2/C2)(−5.0×10−6C)1.0mV2=−4.5×104V
Since the electric potential is the scalar quantity, therefore,
The electric petential at point 'P' due to the two given charges:
VP=V1+V2VP=9.0×104V+(−4.5×104V)VP=4.5×104VVP=V1+V2VP=9.0×104V+(−4.5×104V)VP=4.5×104V
When third charge q3=+2μC=+2.0×10−6Cq3=+2μC=+2.0×10−6C is placed at point 'P', then
The electric potential energy of the charge q3q3 can be expressed as follows:
U3=q3VPU3=q3VP
After plugging in the values, we have:
U3=(+2.0×10−6C)(−4.5
Step-by-step explanation:
hope this helps