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Physics Question on Electromagnetic waves

An electric dipole is formed by two charges +q and -q located in the x-y plane at (0,2) mm and (0,-2) mm as shown in the figure. The electric potential at point P(100, 100) mm due to dipole is V0. Now the charges +q and -q are moved to (-1,2)mm and (1,-2)mm respectively. What is the value of the electric potential at point P due to this new dipole?
An electric dipole is formed by two charges +q and -q located in the x-y plane at 0,2 mm and 0,-2 mm

A

V02\frac{V_0}{2}

B

V04\frac{V_0}{4}

C

V08\frac{V_0}{8}

D

2V02V_0

Answer

V02\frac{V_0}{2}

Explanation

Solution

The correct option is (A):
p1⃗=q×4×10−3  j^\vec{p_1}=q\times 4\times 10^{-3}\,\,\hat{j}

Vat(100,100)mm,V0=Kp1⃗.r1⃗∣r1⃗∣3V_{at}(100,100)mm, V_0=\frac{K\vec{p_1}.\vec{r_1}}{|\vec{r_1}|^3}

V0=9×109×q×[4×10−3j^].(0.1i^+0.1j^)((0.1)2+(0.1)2)3V_0=\frac{9\times10^9\times q\times [4\times 10^{-3}\hat{j}].(0.1\hat{i}+0.1\hat{j})}{(\sqrt{{(0.1)^2}+(0.1)^2})^{3}}

V0=9×109×q× [0.4×10−3](0.1)2×22V_0=\frac{9\times10^9\times q\times\,[0.4\times10^{-3}]}{(0.1)^2\times2\sqrt2}

now +q and -q is moved to (-1,2) mm and (1,-2) mm
p⃗=q[−2i^+4j^]×10−3\vec{p}=q[-2\hat{i}+4\hat{j}]\times10^{-3}

r1⃗=0.1i^+0.1j^\vec{r_1}=0.1\hat{i}+0.1\hat{j}

V=9×109(p⃗,r1⃗)∣r1⃗∣3=9×910×q×[0.4×10−3]((0.1)2+(0.12))3⇒V=v02V=\frac{9\times10^9(\vec{p},\vec{r_1})}{|\vec{r_1}|^3}=\frac{9\times9^{10}\times q\times[0.4\times10^{-3}]}{(\sqrt{{(0.1)^2+(0.1^2)}})^3}\Rightarrow V=\frac{v_0}{2}