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Consider the two charges on the x-axis in the diag

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Answer #1

1)from diagram total electric potential at point P is V=V1+V2

here V1=1/(4\pi \epsilon0)*(q1/r)=9*109*(4.35*10-6/(\sqrt{}((222+642)*10-4)=57.85*103V

V2=1/(4\pi \epsilon0)*(q2/r)=9*109*(6.79*10-6/(\sqrt{}((222+642)*10-4)=90.30*103V

total potential V=57.85*103V+90.30*103V=148.15*103V

2)magnitude of Electric field at origin is E=1/(4\pi \epsilon0)*(q2-q1)/r2

   E=9*109*(6.79-4.35)*10-6/(22*10-2)2=453*103N/c

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