Let the direction of dipole is along the positive direction of x-axis.
Since,
E(r)=4πε0r33(p⋅r^)r^−p .And,
p=25i^r=0.3(cos(15∘)i^+sin(15∘)j^) Now.
\bold p\cdot\hat\bold r=25\cos(15^\circ) Thus,
E(r)=0.339×109(3(25cos(15∘))(cos(15∘)i^+sin(15∘)j^)−25i^⟹E=31012(44.97i^+18.75j^)⟹E=16.24×1012N/C
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